√ Quadratic Equation Solver

Enter a, b, and c from ax² + bx + c = 0 and get the discriminant, both roots, and the vertex — with every step shown so you can follow the working.

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How the quadratic equation solver works

The method, step by step

Step 1: compute the discriminant, D = b² − 4ac. Its sign tells you what kind of roots to expect. Step 2: apply the quadratic formula, x = (−b ± √D) ÷ 2a, once with the plus sign and once with the minus. Step 3: find the vertex at x = −b ÷ 2a and substitute back for its y-value.

Checking your answer

Two useful checks: the roots should add up to −b ÷ a, and their product should equal c ÷ a. For x² − 5x + 6, the roots 2 and 3 add to 5 and multiply to 6 — exactly −b/a and c/a. If your roots fail these checks, a sign was dropped somewhere.

Worked example

x² − 5x + 6 = 0 (a = 1, b = −5, c = 6): D = 25 − 24 = 1, which is positive, so two real roots. x = (5 ± 1) ÷ 2, giving x = 3 and x = 2. The vertex sits at (2.5, −0.25) — the bottom of the upward-opening parabola.

Frequently asked questions

What is the quadratic formula?

For ax² + bx + c = 0, the solutions are x = (−b ± √(b² − 4ac)) / 2a. The ± means there are usually two answers: one using the plus sign and one using the minus sign.

What does the discriminant tell me?

The discriminant D = b² − 4ac decides the kind of answer before you finish solving: D > 0 means two distinct real roots, D = 0 means one repeated root, and D < 0 means no real roots (the parabola never touches the x-axis).

What is the vertex and why does it matter?

The vertex is the turning point of the parabola, at x = −b / 2a. It is the maximum or minimum value of the quadratic — useful for optimisation problems like maximum area or minimum cost.

How do I enter an equation like 2x² = 8x − 6?

Move everything to one side first: 2x² − 8x + 6 = 0, so a = 2, b = −8, c = 6. Watch the signs — a term that crosses the equals sign flips its sign.

Why did I get two answers when my homework expects one?

Most quadratics genuinely have two roots. Check whether the context rules one out — a negative length or a time before the start is often the one to discard.

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