Quadratic Equation Solver

Q: How do I find the discriminant?

D = b² − 4ac. For 2x² + 5x − 3 = 0, D = 5² − 4 · 2 · (−3) = 49.

Q: What if the discriminant is zero?

There is one root, x = −b ÷ 2a. For x² − 4x + 4 = 0 it is x = 2.

Q: What if the discriminant is negative?

There are no real roots: the parabola never crosses the x-axis. The calculator shows the complex roots p ± qi.

Q: How do I check the roots?

Substitute them back, or use Vieta's formulas: the roots add up to −b/a and multiply to c/a.

Q: How do I solve x² − 3x + 2 = 0?

The discriminant is 1, giving roots x₁ = 2 and x₂ = 1. Vieta's formulas confirm them: the roots sum to 3 and multiply to 2.

Q: What does the sign of the discriminant tell me?

Positive means two distinct roots, zero means one repeated root, and negative means no real roots at all — the parabola misses the axis. The calculator also reports the vertex.

Written byOleh Soloninupdated 5 October 2026

Result

Rootsx₁ = 2; x₂ = 1
Discriminant1
Working
D = b² − 4ac = (−3)² − 4 · 1 · 2 = 1 D > 0, so there are two roots. x₁ = (3 + 1) ÷ 2 = 2; x₂ = (3 − 1) ÷ 2 = 1
Check by Vieta's formulasx₁ + x₂ = 3 = −b/a; x₁ · x₂ = 2 = c/a
Vertex of the parabola(1.5, −0.25)

Both roots, and why there are that many

Solves ax² + bx + c = 0 with the discriminant, showing the full working, a check by Vieta's formulas and the vertex of the parabola; complex roots when D < 0.

A quadratic ax² + bx + c = 0 is solved with the discriminant D = b² − 4ac. If D > 0 there are two roots x = (−b ± √D) ÷ 2a; if D = 0, one root x = −b ÷ 2a; if D < 0, no real roots.

For 2x² + 5x − 3 = 0: D = 5² − 4 · 2 · (−3) = 25 + 24 = 49, √D = 7, so x₁ = (−5 + 7) ÷ 4 = 0.5 and x₂ = (−5 − 7) ÷ 4 = −3.

How to use

Rearrange the equation into ax² + bx + c = 0 and enter a, b and c with their signs.
Read the discriminant, the roots and the step-by-step working.
Check against Vieta's formulas: the roots add up to −b/a and multiply to c/a.

Formulas and Vieta's check

Discriminant: D = b² − 4ac. Roots: x = (−b ± √D) ÷ (2a). Vieta's formulas: x₁ + x₂ = −b/a and x₁ · x₂ = c/a, handy for checking an answer or spotting roots mentally when a = 1.

The vertex of y = ax² + bx + c is at x = −b ÷ 2a. When D = 0 the vertex sits on the x-axis and is the only root.

Worked out with this calculator.
Equation	D	Roots
x² − 3x + 2 = 0	1	x₁ = 2; x₂ = 1
2x² + 5x − 3 = 0	49	x₁ = 0.5; x₂ = −3
x² − 4x + 4 = 0	0	x = 2
x² − 3x + 1 = 0	5	x = (3 ± √5) ÷ 2 ≈ 2.618; 0.382
x² + x + 1 = 0	−3	no real roots

What this calculator shows

The roots: one, two or complex
The discriminant and step-by-step working with your coefficients
A check by Vieta's formulas
The vertex of the parabola

What to keep in mind

a cannot be zero, or the equation is linear.
Irrational roots are written with a square root in the working and as decimals in the answer.

FAQs

How do I find the discriminant?

D = b² − 4ac. For 2x² + 5x − 3 = 0, D = 5² − 4 · 2 · (−3) = 49.

What if the discriminant is zero?

There is one root, x = −b ÷ 2a. For x² − 4x + 4 = 0 it is x = 2.

What if the discriminant is negative?

There are no real roots: the parabola never crosses the x-axis. The calculator shows the complex roots p ± qi.

How do I check the roots?

Substitute them back, or use Vieta's formulas: the roots add up to −b/a and multiply to c/a.

How do I solve x² − 3x + 2 = 0?

The discriminant is 1, giving roots x₁ = 2 and x₂ = 1. Vieta's formulas confirm them: the roots sum to 3 and multiply to 2.

What does the sign of the discriminant tell me?

Positive means two distinct roots, zero means one repeated root, and negative means no real roots at all — the parabola misses the axis. The calculator also reports the vertex.

Worked example

An equation that factorises

Input: a = 1, b = −5, c = 6

Output: x₁ = 3; x₂ = 2

Note: D = 25 − 24 = 1, two roots. Vieta: 3 + 2 = 5 = −b/a and 3 · 2 = 6 = c/a. It factorises as (x − 2)(x − 3) = 0.

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Result

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