Quadratic Formula Calculator

Find the roots of any equation ax² + bx + c = 0 and explore the parabola behind it — discriminant analysis, vertex, axis of symmetry, y-intercept, factorisation, an interactive graph, and a full step-by-step derivation for every solution.

Solve ax² + bx + c = 0

ax² + bx + c = 0

Fractions like 5/2, decimals, and negatives are all accepted. a cannot equal zero.

2x² − 7x + 3 = 0

Reading the discriminant

The discriminant Δ = b² − 4ac decides the kind of solutions a quadratic has before you finish solving it. Memorise the three cases — they reappear on every test.

Δ > 0

Two distinct real roots. The parabola crosses the x-axis at two different points.

2x² − 7x + 3 → Δ = 25

Δ = 0

One repeated real root. The parabola only touches the x-axis at the vertex.

x² − 4x + 4 → Δ = 0

Δ < 0

Two complex conjugate roots. The parabola never crosses the x-axis.

x² + 4x + 8 → Δ = −16

Formula reference

Quadratic formula

x = (−b ± √(b² − 4ac)) / (2a)

Solves any ax² + bx + c = 0 with a ≠ 0.

Discriminant

Δ = b² − 4ac

Classifies the roots before you solve.

Vertex (x-coord)

h = −b / (2a)

Lies halfway between the two real roots.

Vertex (y-coord)

k = c − b² / (4a)

Plug h into the original equation, or use this shortcut.

Axis of symmetry

x = −b / (2a)

The same x as the vertex — parabola is mirror-symmetric about it.

Y-intercept

y = c

Set x = 0 in ax² + bx + c.

Sum of roots

x₁ + x₂ = −b / a

Vieta's first formula.

Product of roots

x₁ · x₂ = c / a

Vieta's second formula — useful sanity check after solving.

Derivation of the quadratic formula

The quadratic formula is not magic — it is what you get when you complete the square on the general equation ax² + bx + c = 0. Reading through the derivation once makes the formula far easier to remember and to apply correctly under exam pressure.

  1. 1

    Start with the general quadratic

    Any second-degree polynomial equation can be written in standard form with three real coefficients. We assume a is non-zero — otherwise the equation collapses to a linear one.

    ax² + bx + c = 0

  2. 2

    Divide every term by a

    Dividing by a leaves a monic quadratic (leading coefficient 1). This is legal because a ≠ 0 and it sets up the completing-the-square move that follows.

    x² + (b/a) · x + (c/a) = 0

  3. 3

    Move the constant to the other side

    Subtract c/a from both sides so the variable terms are isolated on the left. This pure-quadratic-plus-linear form is ready to be completed into a perfect square.

    x² + (b/a) · x = − c/a

  4. 4

    Add (b/2a)² to both sides

    Half of the linear coefficient is b/(2a). Squaring it gives b²/(4a²). Adding this constant to both sides preserves the equation and makes the left a perfect square trinomial.

    x² + (b/a) · x + (b/2a)² = (b/2a)² − c/a

  5. 5

    Factor the left as a perfect square

    The left-hand trinomial now equals (x + b/(2a))² exactly. The right side is a single fraction — combine the terms over the common denominator 4a².

    (x + b/(2a))² = (b² − 4ac) / (4a²)

  6. 6

    Take square roots of both sides

    Square-rooting kills the square on the left and introduces a ± on the right — because both the positive and negative root satisfy the original equation. The denominator 4a² has root 2a (we take the positive root; the sign is absorbed by the ±).

    x + b/(2a) = ± √(b² − 4ac) / (2a)

  7. 7

    Isolate x — the quadratic formula appears

    Subtract b/(2a) from both sides. Combine the two fractions over the common denominator 2a and you are left with the formula every algebra textbook prints inside a box.

    x = (−b ± √(b² − 4ac)) / (2a)

Four ways to solve a quadratic

MethodWhen it works bestLimitation
Quadratic formulaAlways works — first choice when coefficients are messy decimals, fractions, or large integers.Slightly slower to write out; easy to forget the ± or misread b².
FactoringInteger roots and small integer coefficients — fastest when (px + q)(rx + s) jumps out at a glance.Fails whenever the roots are irrational, complex, or have ugly fractions.
Completing the squareBuilds the derivation of the formula and is invaluable for converting to vertex form.More algebra steps than the formula; arithmetic mistakes compound quickly.
GraphingGives a quick visual estimate and confirms the kind of roots — great as a sanity check.Only approximate without algebra; misreads roots that are close together or off-screen.

What Is A Quadratic Formula Calculator?

A quadratic formula calculator solves any equation of the form ax² + bx + c = 0 for the unknown x. You enter the three coefficients a, b, and c; the calculator computes the discriminant, classifies the roots, returns both solutions (real or complex), graphs the underlying parabola, and walks through every line of the derivation — substitution, square root, denominator, positive branch, negative branch. It is a study tool, a homework checker, and a fast scratchpad all in one.

This tool accepts integers, decimals, simple fractions like 5/2, negative values, and the constant π. It detects integer factorisations automatically, draws the parabola with the vertex and axis of symmetry marked, and reports every property a textbook question can ask — vertex coordinates, y-intercept, opening direction, minimum or maximum value, domain and range. It pairs naturally with the scientific calculator, the root calculator, and the exponent calculator.

Understanding The Quadratic Equation

Three coefficients fix the curve

Every quadratic is determined by exactly three numbers: a (the curvature, must be non-zero), b (the slant), and c (the height at x = 0). Pick any three real numbers with a ≠ 0 and you have specified a unique parabola — there is no other curve that fits.

Roots are x-intercepts

Setting ax² + bx + c = 0 is the same as asking where the parabola meets the horizontal axis. Two crossings → two real roots, one touch → a repeated root, no contact → complex conjugate roots. The discriminant tells you which case you are in before you even open your calculator.

Symmetry is everywhere

A parabola is perfectly mirror-symmetric about its vertical axis of symmetry x = −b/(2a). The two real roots sit equidistant from that line, which is why their sum equals −b/a (Vieta's first formula) and the midpoint is the vertex's x-coordinate.

The vertex is the optimum

Because parabolas open straight up or straight down, the vertex is either the minimum (when a > 0) or the maximum (when a < 0). Optimisation problems in physics, economics, and engineering reduce to finding this single point — and the quadratic formula does it implicitly through the axis of symmetry.

6 Ways To Use This Quadratic Calculator

1

Check homework instantly

Type the coefficients, hit Calculate, and compare against your own work. The step-by-step panel makes it obvious where an arithmetic slip happened — usually in the discriminant or the final fraction.

2

Classify roots before solving

The discriminant badge tells you whether the roots are real, repeated, or complex before you commit to a method. Saves time on multiple-choice questions where the type of root is the answer.

3

Visualise the parabola

The interactive graph plots y = ax² + bx + c with the vertex, axis of symmetry, and intercepts marked. Useful for sketching answers on graph paper or recognising the curve from a word problem.

4

Find vertex and optimum

Looking for the maximum height of a projectile or the minimum cost of a function? The vertex card returns h, k, the axis of symmetry, and labels k as the minimum or maximum based on the sign of a.

5

Spot integer factorisations

If the equation factors cleanly, the calculator finds and prints the factored form like (2x − 1)(x − 3). When no integer factorisation exists, it says so directly — no guessing.

6

Practice complex roots

Try the x² + 4x + 8 preset to see the calculator produce −2 ± 2i with a full conjugate-root derivation. Great for an Algebra II or Pre-Calc unit on the imaginary unit.

Best Practices For Solving Quadratics

Always rewrite the equation in standard form ax² + bx + c = 0 before reading off the coefficients. Equations are often presented as 3x² = 7x − 2 or x(x − 5) = 4; until you collect all terms on the same side, the values of b and c you plug in will be wrong. Watch the signs carefully — a stray minus sign on b² is the single most common error students make on the discriminant.

Compute the discriminant first, then take its square root only once. Re-evaluating b² − 4ac twice (once for each branch of the ±) doubles the chance of making an arithmetic slip. The disciplined sequence is: discriminant → √Δ → denominator → x₁ → x₂. The calculator follows exactly this order, and you should too on paper.

Verify your answers with Vieta's formulas. If your two roots sum to −b/a and multiply to c/a, you almost certainly have the right answer. This 10-second sanity check has saved more exam marks than any other trick. The summary table on this calculator surfaces the values needed for the check.

Why Quadratic Equations Matter

Projectile motion

The trajectory of any object launched under gravity is a parabola in the height-vs-time graph: h(t) = −½ g t² + v₀ t + h₀. Solving for when h(t) = 0 (impact time) or the vertex (peak height) is a quadratic problem in disguise — every kinematics chapter relies on it.

Economics and optimisation

Revenue, cost, and profit curves are routinely modelled as quadratics. Maximum revenue at the vertex, break-even points at the roots, and marginal cost from the derivative all come from the same equation. Even simple supply-and-demand models bottom out as quadratic intersections.

Computer graphics and games

Bezier curves, parabolic arcs, and lens equations in rendering pipelines reduce to quadratics. Ray–sphere intersection — the workhorse of ray tracing — is solved by the quadratic formula on every frame. The same calculation drives collision detection between circles and lines in 2D engines.

Engineering and architecture

Hanging cables under uniform load (suspension bridges, parabolic arches), the cross-section of a satellite dish, and the reflector behind a car headlight are all parabolas. Their structural and optical properties are derived from the same ax² + bx + c that this calculator solves.

Where Quadratic Problems Get Tricky

Hidden quadratics

Equations like (x − 3)(x + 7) = 5 or 1/x + 1/(x + 2) = 3 are quadratics in disguise. Expand, clear denominators, and collect — only then can you read off a, b, and c. This calculator only accepts the standard form, so do that algebra first.

Quadratic in something else

x⁴ − 5x² + 6 = 0 is a quadratic in the substitution u = x². Solve for u with this calculator, then take square roots to recover x. The same trick works for x⁶ − 7x³ + 12, eˣ-based equations, and many trig identities.

Coefficients with units

If your coefficients carry physical units, the discriminant and roots inherit them. Always solve symbolically first, then plug in numbers at the end — that catches dimensional mistakes (m vs cm, s vs ms) before they propagate.

Repeated roots in disguise

When Δ is a small floating-point number like 1e-14, the equation has a repeated root in exact arithmetic but the formula returns two roots that differ by 1e-7. The calculator snaps such near-zero discriminants to zero and reports the repeated case correctly.

Core Quadratic Formulas

Standard form

ax² + bx + c = 0

Every quadratic problem starts here.

Quadratic formula

x = (−b ± √(b² − 4ac)) / (2a)

Universal solution method.

Discriminant

Δ = b² − 4ac

Classifies the roots.

Vertex x-coordinate

h = −b / (2a)

Axis of symmetry of the parabola.

Vertex y-coordinate

k = c − b² / (4a)

Minimum (a > 0) or maximum (a < 0) value.

Vertex form

y = a (x − h)² + k

Useful for transformations and graphing.

Sum of roots

x₁ + x₂ = −b / a

Vieta's first formula.

Product of roots

x₁ · x₂ = c / a

Vieta's second formula.

Factored form

a (x − x₁)(x − x₂)

Equivalent representation when roots are known.

Y-intercept

y(0) = c

Where the parabola meets the y-axis.

Average of roots

(x₁ + x₂) / 2 = h

Midpoint of the roots is the vertex's x.

Distance between roots

|x₁ − x₂| = √Δ / |a|

Width of the gap between the two zeros.

Derivation Of The Quadratic Formula

The quadratic formula is not a magic incantation. Reading through the derivation once makes the formula impossible to forget — it is the result of dividing through by a, completing the square, and taking the square root of both sides. The line-by-line working below is the same proof that appears in every algebra textbook, rewritten with consistent spacing.

ax² + bx + c = 0

Multiply by 1/a

x² + (b/a)x + c/a = 0

x² + (b/a)x = − c/a

From this point, it is possible to complete the square using the identity:

x² + bx + c = (x − h)² + k

Continuing the derivation using this relationship:

x² + (b/a)x + (b/2a)² = − c/a + (b/2a)²

Simplify

(x + b/2a)² = − 4ac/4a² + b²/4a² = (b² − 4ac) / 4a²

Square root both sides

x + b/2a = ± √(b² − 4ac) / 2a

Solve for x

x = (−b ± √(b² − 4ac)) / 2a

Two observations from the derivation: the ± exists because the square root of any positive number has two values, and the discriminant b² − 4ac sits naturally inside the radical — so its sign decides everything about the kind of roots before you take the square root at all. The same line of reasoning produces the vertex form y = a(x − h)² + k if you stop at step 5 and rearrange instead of square-rooting.

Common Quadratic Mistakes

  1. 1

    Skipping the standard-form rewrite

    Students plug coefficients in from an unsimplified equation like 3x² + 5 = 7x and end up with the wrong b. Always move every term to the left side first so the right side is exactly 0.

  2. 2

    Sign mistake on b

    When b is negative, the formula uses −b — which becomes positive. −(−7) is +7, not −7. The single most common sign error on quadratic exams comes from forgetting this double negative.

  3. 3

    Computing b² with the sign

    (−7)² = 49, not −49. b is squared, so the sign of b never reaches the discriminant. Wrap negatives in parentheses before squaring on a calculator.

  4. 4

    Forgetting the ±

    Returning only one root is the second most common slip. The ± is the whole point of the formula — write both branches down explicitly even if one looks ugly.

  5. 5

    Dividing instead of multiplying 2a

    The denominator is 2a, not 2/a or a/2. For a = 3 the denominator is 6, not 1.5 or 0.667. A wrong denominator throws both roots off.

  6. 6

    Confusing factored form with roots

    (2x − 1)(x − 3) = 0 has roots x = 1/2 and x = 3, not x = 1 and x = 3. Set each factor to zero and solve — don't just read the constants.

Three Worked Examples

Example 1 · Two real roots

Solve 2x² − 7x + 3 = 0

a = 2, b = −7, c = 3. Discriminant Δ = (−7)² − 4 · 2 · 3 = 49 − 24 = 25, so √Δ = 5. The denominator 2a = 4. The roots come out as x = (7 + 5) / 4 = 3 and x = (7 − 5) / 4 = 1/2 = 0.5. Verify with Vieta: 3 + 0.5 = 3.5 = 7/2 = −b/a and 3 · 0.5 = 1.5 = 3/2 = c/a. ✓

Example 2 · Repeated root

Solve x² − 4x + 4 = 0

a = 1, b = −4, c = 4. Discriminant Δ = 16 − 16 = 0 → repeated root. x = −b/(2a) = 4/2 = 2. Equivalent factored form: (x − 2)². The parabola y = x² − 4x + 4 just touches the x-axis at x = 2 and never crosses it. Vertex (2, 0).

Example 3 · Complex roots

Solve x² + 4x + 8 = 0

a = 1, b = 4, c = 8. Discriminant Δ = 16 − 32 = −16, so √Δ = 4i. The roots are x = (−4 ± 4i) / 2 = −2 ± 2i. Complex conjugates as expected from real coefficients. The parabola y = x² + 4x + 8 lies entirely above the x-axis (vertex (−2, 4)), so it never intersects y = 0.

Real-Life Applications

Ballistics and sports

Basketball arcs, football punts, golf-ball flight paths, and artillery trajectories are all parabolas through still air. The roots are the launch and landing points; the vertex is the apex; the discriminant decides whether a shot can clear a given height.

Profit maximisation

If revenue is linear and cost is linear, profit is a quadratic in production volume. The vertex gives the optimal quantity to make, and the roots give the break-even points where profit returns to zero. Every Econ 101 textbook leans on this.

Chemistry equilibria

Acid-base equilibria, Ksp calculations, and ICE-table problems frequently produce a quadratic in the unknown concentration. Solving for the positive root of that quadratic gives the equilibrium concentration; the negative root is discarded as unphysical.

Computer graphics

Ray–sphere intersection — used in every ray tracer and most rasterised lighting models — boils down to solving (D · D)t² + 2(O · D)t + (O · O − r²) = 0. The discriminant says whether the ray hits the sphere; the smaller positive root is the first contact.

Tips For Solving Quadratics Faster

  • Look for the trivial factorisation first. If a, b, and c are small integers and the equation is monic, try (x − p)(x − q) where pq = c and p + q = −b. Many exam quadratics fall to this in seconds.
  • Spot perfect squares. If b² = 4ac, the equation is a perfect square trinomial and the root is just −b/(2a). No formula needed.
  • Use Vieta as a check, not a solve. Vieta gives two equations in two unknowns; pairing them with the quadratic formula doesn't speed up solving but does verify it in seconds.
  • Clear fractions before plugging in. An equation like ½ x² − x − 3 = 0 is easier to work with after multiplying through by 2 → x² − 2x − 6 = 0. Same roots, smaller numbers.
  • Sketch the parabola when stuck. A rough graph reveals quickly whether the discriminant is positive (two crossings), zero (a touch), or negative (no contact). The visual sanity check kills half of all sign mistakes.

Methodology you can verify

Every result is computed from the canonical quadratic formula in double-precision floating point. The discriminant is calculated as b² − 4ac, the square root is taken from Math.sqrt, and the denominator is 2a. Values within 1e-9 of a clean integer, half, or quarter are snapped back to the exact rational — so an answer of 0.49999999 is reported as 0.5. Integer factorisations are found by exhaustive search over the divisors of a · c, and complex roots are returned as a real part ± an imaginary part. Read more on the methodology and editorial policy pages.

Frequently Asked Questions

A quadratic equation is any polynomial equation of the second degree in a single variable — that is, any equation that can be rewritten in the standard form ax² + bx + c = 0 with a ≠ 0. The numbers a, b, and c are the coefficients (a is the quadratic coefficient, b the linear coefficient, and c the constant term). Quadratic equations describe parabolas, projectile motion under gravity, the area of any rectangle whose side length is variable, the height of cables under uniform load, and many optimisation problems in economics and engineering — virtually any situation where a quantity depends on the square of a variable.

The quadratic formula gives every solution of ax² + bx + c = 0 in a single expression: x = (−b ± √(b² − 4ac)) / (2a). The ± produces two roots — one from the + branch and one from the − branch — which are the same value when the discriminant b² − 4ac equals zero, and complex conjugates when it is negative. The formula is derived from the standard equation by dividing through by a and completing the square, and it works for any real a, b, c with a ≠ 0.

The discriminant is the quantity Δ = b² − 4ac that appears under the square root inside the quadratic formula. Its sign decides how many real solutions the equation has: Δ > 0 means two distinct real roots, Δ = 0 means one repeated real root, and Δ < 0 means two complex conjugate roots (no real solutions). The discriminant is also the square of the gap between the two roots (multiplied by a²) — large positive discriminants correspond to roots that are spread far apart, small positive discriminants to roots that are close together.

A quadratic is a second-degree polynomial, and the Fundamental Theorem of Algebra guarantees exactly n solutions for any nth-degree polynomial in the complex numbers (counted with multiplicity). For a degree-2 polynomial that means two roots — they can be two distinct real numbers (the parabola crosses the x-axis twice), one repeated real number (the parabola just touches the x-axis), or two complex conjugates (the parabola sits entirely above or below the x-axis). The ± in the quadratic formula reflects the two roots geometrically.

When the discriminant b² − 4ac is negative, the square root in the quadratic formula becomes the square root of a negative number — which is not a real number. Factoring out √(−1) = i (the imaginary unit) lets you express the two roots as a complex conjugate pair x = −b/(2a) ± (√|Δ|/(2a)) · i. The parabola never touches the x-axis because both roots are off the real number line. This is perfectly legal mathematics, and the calculator returns both complex roots with their real and imaginary parts spelled out.

Yes. Any quadratic with real coefficients and a negative discriminant has two complex roots, and they always come as a conjugate pair — same real part, opposite imaginary part. For example, x² + 4x + 8 = 0 has roots −2 ± 2i. Conjugate roots are not a special case to memorise; they are a direct consequence of the formula, because the ± of the imaginary square root flips the sign of the imaginary part while keeping the real part fixed. Complex roots are central to electrical engineering (impedance), quantum mechanics (wavefunctions), and signal processing (Fourier theory).

Three quick checks: (1) the discriminant must be a perfect square (so the square root is rational); (2) both roots from the quadratic formula must come out as rational numbers (no irrational tails); and (3) for integer coefficients, you need integer p, q, r, s with pr = a, qs = c, and ps + qr = b. This calculator searches the divisors of a and c automatically — when an integer factorisation exists it returns the factored form like (2x − 1)(x − 3); otherwise it tells you that no simple integer factorisation exists, and you fall back on the quadratic formula or completing the square.

The vertex is the single turning point of a parabola — the lowest point when the parabola opens upward (a > 0), the highest point when it opens downward (a < 0). For y = ax² + bx + c, the vertex sits at (h, k) where h = −b/(2a) and k = c − b²/(4a). The vertex is the optimum value of any quadratic objective function — maximum revenue, minimum cost, peak height of a projectile, lowest point of a sagging cable — and finding it is half of all real-world quadratic problems.

The coefficient a controls both the direction the parabola opens and how steeply it curves. If a > 0 the parabola opens upward (smile) and the vertex is a minimum; if a < 0 it opens downward (frown) and the vertex is a maximum. Larger |a| makes the parabola narrower (steeper sides); smaller |a| makes it wider. Crucially, a cannot equal 0 — if it did, the entire x² term would vanish and the equation would degenerate to the linear bx + c = 0, which has at most one solution and no parabola at all.

The axis of symmetry is the vertical line x = −b/(2a) that splits a parabola into two mirror-image halves. For any point on one side of the axis, there is a matching point at the same height on the other side. The two real roots of a quadratic are equidistant from this axis (which is why their sum is twice the vertex's x-coordinate, equal to −b/a — Vieta's formula). The axis of symmetry passes through the vertex and runs perpendicular to the x-axis.

All three methods produce the same answers — they are different algorithms for the same problem. Factoring is fastest when integer factorisation jumps out at you (small integer coefficients, rational roots) but fails for irrational or complex roots. Completing the square turns the equation into a perfect-square form (x − h)² = k that solves directly; it is also the proof method behind the quadratic formula itself and the standard way to convert to vertex form. The quadratic formula always works, requires no insight, and is the safest choice on a timed exam. Graphing only gives approximate answers but is a quick sanity check.

Yes. The quadratic formula doesn't care whether the coefficients are integers, fractions, decimals, irrationals, or negatives — any real number is acceptable as long as a ≠ 0. This calculator accepts simple fractions like 5/2 or 3/4, decimals like 0.75 or 2.5, the constant π, and negative values. For neatness when factoring, you can often multiply through by the common denominator to clear fractions before plugging in — the roots are unaffected by scaling the entire equation by a non-zero constant.