Quadratic Equation Solver – Roots, Discriminant and Vertex
Enter a, b and c to solve ax² + bx + c = 0. You get the roots — real or complex — the discriminant, the vertex of the parabola and the quadratic-formula steps.
Use x = (−b ± √(b² − 4ac)) ÷ 2a. For x² − 3x − 10 = 0: a = 1, b = −3, c = −10, the discriminant is 9 + 40 = 49, so x = (3 ± 7) ÷ 2, giving x = 5 and x = −2.
How to use Quadratic Equation Solver
- Rewrite the equation in the form ax² + bx + c = 0.
- Enter a, b and c, including negative signs.
- Read the solutions.
- Check the discriminant to see how many real roots there are.
- Use the vertex and axis of symmetry to sketch the parabola.
The quadratic formula
Every quadratic equation ax² + bx + c = 0 with a ≠ 0 can be solved with:
x = (−b ± √(b² − 4ac)) ÷ 2a
Example: x² − 3x − 10 = 0. Here a = 1, b = −3, c = −10. The discriminant is (−3)² − 4 · 1 · (−10) = 9 + 40 = 49, and √49 = 7. So x = (3 + 7) ÷ 2 = 5 or x = (3 − 7) ÷ 2 = −2. Check: 5² − 3·5 − 10 = 0 ✓.
The discriminant
The value D = b² − 4ac tells you what kind of solutions to expect before you finish:
| Discriminant | Solutions | Graph |
|---|---|---|
| D > 0 | Two different real roots | Parabola crosses the x-axis twice |
| D = 0 | One repeated real root | Parabola touches the x-axis |
| D < 0 | Two complex roots | Parabola does not reach the x-axis |
When D < 0, the roots are complex numbers of the form p ± qi, where i = √−1. For x² + 2x + 5 = 0, D = −16 and x = −1 ± 2i.
Vertex and axis of symmetry
The parabola y = ax² + bx + c has its turning point, the vertex, at x = −b ÷ 2a. Put that x back into the equation to get the y-value. For x² − 3x − 10 the vertex is at (1.5, −12.25). If a > 0 the parabola opens upward and the vertex is a minimum; if a < 0 it opens downward and the vertex is a maximum.
Other ways to solve
- Factoring: x² − 3x − 10 = (x − 5)(x + 2), so x = 5 or −2. Fast when it works.
- Completing the square: rewrite as (x − 1.5)² = 12.25 and take square roots.
- Graphing: read where the parabola crosses the x-axis.
The formula always works, which is why the solver uses it. If a = 0 the equation is linear and the solver returns x = −c ÷ b instead.
Where quadratics appear
Projectile height, areas, profit models and braking distances often lead to quadratic equations. For the square roots inside the formula, see the Square Root Calculator calculator; for straight lines, use the Slope Calculator.
You may also find these free tools useful: GPA Calculator, LCM & GCF Calculator and Slope Calculator.
Frequently asked questions
How do you solve a quadratic equation?
Put it in the form ax² + bx + c = 0 and use x = (−b ± √(b² − 4ac)) ÷ 2a, or factor it if possible.
What does the discriminant tell you?
If b² − 4ac is positive there are two real roots, if zero one repeated root, and if negative two complex roots.
What if the discriminant is negative?
The equation has no real solutions; its roots are complex numbers such as −1 ± 2i.
How do I find the vertex of a parabola?
Its x-coordinate is −b ÷ 2a. Substitute it into the equation to find the y-coordinate.
What if a is zero?
Then the equation is linear, bx + c = 0, with the single solution x = −c ÷ b.
Last updated: