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Quadratic Equation Solver – Solve ax² + bx + c = 0

Enter the coefficients a, b and c to solve any quadratic equation. You’ll get the real or complex roots, the discriminant, the vertex of the parabola and the factored form, with each step of the quadratic formula shown.

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How to use the quadratic equation solver

  1. Rearrange your equation so one side is zero: ax² + bx + c = 0.
  2. Enter a, b and c — including minus signs and zeros for missing terms.
  3. Read the roots, the discriminant, the vertex and the factored form.

For example, 2x² = 8 becomes 2x² + 0x − 8 = 0, so a = 2, b = 0 and c = −8.

The quadratic formula

x = (−b ± √(b² − 4ac)) ÷ 2a
Discriminant D = b² − 4ac
Vertex = (−b ÷ 2a, c − b² ÷ 4a)

Worked example: x² − 3x − 10 = 0

  1. a = 1, b = −3, c = −10
  2. D = (−3)² − 4 × 1 × (−10) = 9 + 40 = 49
  3. √D = 7
  4. x = (3 + 7) ÷ 2 = 5 and x = (3 − 7) ÷ 2 = −2
  5. Factored form: (x − 5)(x + 2) = 0
  6. Vertex: x = 3 ÷ 2 = 1.5, y = −10 − 9 ÷ 4 = −12.25

Worked example with complex roots: x² + 2x + 5 = 0

D = 4 − 20 = −16. The roots are x = (−2 ± 4i) ÷ 2 = −1 ± 2i.

Other ways to solve quadratics

  • Factoring — find two numbers that multiply to ac and add to b. Quick when the roots are whole numbers.
  • Completing the square — rewrite as a(x − h)² + k = 0. This is how the quadratic formula is derived, and it gives the vertex directly.
  • Graphing — the roots are where the parabola crosses the x-axis.

The formula always works, which is why the solver uses it. Behind the scenes it uses a numerically stable version that avoids rounding errors when b is much larger than a and c.

Where quadratics appear

  • Physics: the height of a thrown ball, h = −4.9t² + vt + h₀, is quadratic in time.
  • Area problems: a rectangle whose length is 3 m more than its width and whose area is 40 m² gives w² + 3w − 40 = 0, so w = 5 m.
  • Business: profit often rises and then falls with price, giving a parabola whose vertex is the best price.
  • Engineering and design: parabolic shapes appear in satellite dishes, bridges and car headlights.

Reading the graph

The sign of a tells you the shape: a positive a opens upwards like a cup, a negative a opens downwards like a hill. The vertex is the lowest or highest point, and the roots are where the curve meets the x-axis. If the discriminant is negative, the curve sits entirely above or below the axis, which is why there are no real roots.

Checking your answer

Substitute each root back into the equation. For x = 5 in x² − 3x − 10: 25 − 15 − 10 = 0 ✓. You can also check that the roots add up to −b ÷ a and multiply to c ÷ a: 5 + (−2) = 3 and 5 × (−2) = −10 ✓.

For other calculations, use the scientific calculator, or explore powers and roots with the exponent calculator.

Frequently asked questions

What is the quadratic formula?

x = (−b ± √(b² − 4ac)) ÷ 2a. It gives both solutions of ax² + bx + c = 0 for any a that isn’t zero.

How do I solve x² − 3x − 10 = 0?

Here a = 1, b = −3, c = −10. The discriminant is 9 + 40 = 49, so √49 = 7 and x = (3 ± 7) ÷ 2, giving x = 5 or x = −2. In factored form: (x − 5)(x + 2) = 0.

What does the discriminant tell me?

D = b² − 4ac. If D > 0 there are two different real roots; if D = 0 there is one repeated real root; if D < 0 there are no real roots, only two complex ones.

What are complex roots?

When the discriminant is negative, the roots involve i, the square root of −1. For x² + 2x + 5 = 0, x = −1 ± 2i. The parabola never crosses the x-axis.

What is the vertex of a parabola?

The turning point of the graph y = ax² + bx + c, at x = −b ÷ 2a. For x² − 3x − 10 it is at (1.5, −12.25). If a is positive the vertex is the lowest point; if negative, the highest.

What if a is 0?

Then the equation is linear, bx + c = 0, with the single solution x = −c ÷ b. The solver asks for a non-zero a.

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