Quadratic Equation Solver
Enter a, b and c to solve ax² + bx + c = 0. You get exact roots in simplified radical form (or complex roots), the discriminant, vertex, axis of symmetry, factored and vertex forms, a step-by-step quadratic formula solution and a graph of the parabola.
Enter whole numbers, decimals or fractions such as 3/4 or -0.5. Leave b or c empty for 0.
Graph of the parabola
Step-by-step solution
How to use the Quadratic Equation Solver
- Rewrite your equation so one side is zero, in the form ax² + bx + c = 0.
- Type a, b and c. Use a minus sign for negative values, and decimals or fractions are fine. A missing term is 0.
- Press Solve (results also update as you type).
- Read the roots in exact form and as decimals, plus the discriminant, vertex and factored form.
- Check the graph and steps: the parabola shows where it crosses the x-axis, and the steps show every stage of the quadratic formula.
Quadratic formula and key facts
| Quantity | Formula | Meaning |
|---|---|---|
| Roots | x = (−b ± √(b² − 4ac)) / 2a | Where the parabola crosses the x-axis |
| Discriminant | D = b² − 4ac | D > 0: two real roots · D = 0: one double root · D < 0: complex roots |
| Sum of roots | x₁ + x₂ = −b/a | Vieta's formula, a quick check |
| Product of roots | x₁ · x₂ = c/a | Vieta's formula |
| Vertex | (−b/2a, c − b²/4a) | Minimum if a > 0, maximum if a < 0 |
| Axis of symmetry | x = −b/2a | The parabola is a mirror image on each side |
| Vertex form | y = a(x − h)² + k | Completing the square |
Worked example with an irrational answer
Solve 2x² + 4x − 3 = 0. Here a = 2, b = 4 and c = −3.
- Discriminant: D = 4² − 4·2·(−3) = 16 + 24 = 40. It is positive but not a perfect square, so there are two irrational real roots.
- Formula: x = (−4 ± √40) / 4.
- Simplify the root: √40 = √(4·10) = 2√10, so x = (−4 ± 2√10) / 4.
- Cancel the common factor 2: x = (−2 ± √10) / 2, which is x ≈ 0.581139 or x ≈ −2.581139.
- Check with Vieta: the sum is −2 = −b/a and the product is (4 − 10)/4 = −1.5 = c/a.
Examples of each type of root
| Equation | D | Roots | Factored form |
|---|---|---|---|
| x² − 3x + 2 = 0 | 1 | 1 and 2 | (x − 1)(x − 2) |
| 6x² − 5x − 6 = 0 | 169 | −2/3 and 3/2 | (3x + 2)(2x − 3) |
| x² − 6x + 9 = 0 | 0 | 3 (double root) | (x − 3)² |
| 2x² + 4x − 3 = 0 | 40 | (−2 ± √10)/2 | 2(x − 0.581139)(x + 2.581139) |
| x² + 2x + 5 = 0 | −16 | −1 ± 2i | Not factorable over the reals |
Frequently asked questions
What is the quadratic formula?
For ax² + bx + c = 0 with a ≠ 0, the solutions are x = (−b ± √(b² − 4ac)) / 2a. The ± gives the two roots, one using + and one using −.
What does the discriminant tell you?
The discriminant D = b² − 4ac decides the type of roots. If D > 0 there are two different real roots, if D = 0 there is one repeated real root, and if D < 0 there are two complex conjugate roots and the parabola does not cross the x-axis.
How do I simplify the square root in the answer?
Take out the largest perfect square factor. For example √40 = √(4 × 10) = 2√10, so (−4 ± √40)/4 simplifies to (−2 ± √10)/2 after dividing everything by the common factor 2.
How do you find the vertex of a parabola?
The vertex x-coordinate is h = −b/2a and the y-coordinate is k = c − b²/4a, which is the same as substituting h back into the equation. The vertex is a minimum when a > 0 and a maximum when a < 0.
When can a quadratic be factored?
A quadratic with whole-number coefficients factors into whole-number brackets exactly when its discriminant is a perfect square. For 6x² − 5x − 6, D = 169 = 13², so it factors as (3x + 2)(2x − 3).
What happens if a = 0?
Then there is no x² term and the equation is linear, bx + c = 0, with the single solution x = −c/b. The solver detects this and solves it as a linear equation.
Can I enter fractions or decimals?
Yes. Values like 1/2, −3/4 or 0.25 are converted to exact fractions, so the roots, vertex and factored form stay exact instead of being rounded.