Any equation that can be rearranged to ax² + bx + c = 0 with a ≠ 0 is quadratic. The discriminant decides the type of solution before you compute it.
Reading the discriminant
- D > 0: two different real roots; the parabola crosses the x-axis twice.
- D = 0: one repeated root; the vertex touches the axis.
- D < 0: two complex conjugate roots, shown as real part ± imaginary part.
Check your answer
The roots always satisfy x₁ + x₂ = −b/a and x₁ × x₂ = c/a.
Inputs explained
| Field | What to enter |
|---|---|
| a (x² coefficient) | A number |
| b (x coefficient) | A number |
| c (constant) | A number |
What each result means
- Discriminant (b² − 4ac) = [b (x coefficient)]^2 – 4×[a (x² coefficient)]×[c (constant)]
- Vertex x = -[b (x coefficient)] ÷ (2×[a (x² coefficient)])
Example values
How the main result changes when a changes and everything else stays at the example values above:
| a (x² coefficient) | Discriminant (b² − 4ac) |
|---|---|
| 0.5 | 29 |
| 0.75 | 39 |
| 1 | 49 |
| 1.25 | 59 |
| 1.5 | 69 |
| 2 | 89 |
Formula
x = (−b ± √(b² − 4ac)) ÷ 2aDiscriminant D = b² − 4acWorked example
Solve x² − 3x − 10 = 0 (a = 1, b = −3, c = −10).
D = 9 + 40 = 49, so √D = 7.
x = (3 ± 7) ÷ 2, giving x₁ = 5 and x₂ = −2. Vertex at x = 1.5.
Key terms
- Discriminant
- b² − 4ac, which determines the number and type of roots.
- Root
- A value of x that makes the equation equal zero.
- Vertex
- The turning point of the parabola, at x = −b/2a.
Frequently asked questions
What if a = 0?
The equation is linear, not quadratic; solve bx + c = 0 instead.
How are complex roots shown?
As a real part plus or minus an imaginary part.
Can I factor instead?
If the discriminant is a perfect square, the equation factors neatly over whole numbers.
What formula does the Quadratic Equation Solver use?
x = (−b ± √(b² − 4ac)) ÷ 2a.