Quadratic equation solver

Enter the coefficients a, b and c: the discriminant and the roots show, exact and as decimals, with the parabola and its vertex.

Checked on thousands of cases
Updated November 7, 2026

x² − 5x + 6 = 0

Roots · Two real roots

Discriminant: Δ = b² − 4ac = 1
Exact form: x = 2, x = 3
Vertex of the parabola: (2.5, -0.25)

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How it works

Formula: x = (−b ± √Δ) ÷ 2a, with Δ = b² − 4ac

First compute the discriminant Δ = b² − 4ac. If it is positive, there are two roots: x = (−b ± √Δ) ÷ 2a. If it is zero, one double root: x = −b ÷ 2a. If it is negative, no real root, but two complex roots. The roots are where the parabola y = ax² + bx + c crosses the horizontal axis; its vertex is at x = −b ÷ 2a.

Worked examples

x² − 5x + 6 = 0
Δ = 25 − 24 = 1: two roots, x = 2 and x = 3.
x² − 2x − 1 = 0
Δ = 8: x = 1 ± √2, that is -0.414 and 2.414.
x² + 2x + 5 = 0
Δ = −16: no real root, but x = −1 ± 2i.

Common mistakes

Forgetting the minus sign in front of b.
With b = −5, −b is 5: x = (5 ± 1) ÷ 2.
Dividing only √Δ by 2a.
The whole numerator is divided: (−b ± √Δ) ÷ 2a.
Writing b² = −25 when b = −5.
A square is never negative: (−5)² = 25.

Sample equations and their roots

Sample equations and their roots
Equation	Δ	Roots
x² − 5x + 6 = 0	1	2, 3
x² − 4 = 0	16	−2, 2
x² + 2x + 1 = 0	0	−1
2x² − 3x − 2 = 0	25	−1/2, 2
x² − 2x − 1 = 0	8	1 ± √2
x² − 2 = 0	8	±√2
x² + 2x + 5 = 0	-16	no real root (−1 ± 2i)
x² + 1 = 0	-4	no real root (±i)
3x² − 6x + 3 = 0	0	1
x² − 7x + 12 = 0	1	3, 4

Frequently asked questions

What does a negative discriminant mean?

The parabola doesn't cross the horizontal axis: no real root, two complex conjugate roots.

What if a = 0?

It's no longer quadratic: bx + c = 0 gives x = −c ÷ b.

What are the sum and product of the roots?

Sum = −b ÷ a, product = c ÷ a: for x² − 5x + 6, 2 + 3 = 5 and 2 × 3 = 6.