Quadratic Formula Calculator
Solve ax²+bx+c=0 for real or complex roots; also shows vertex, discriminant and axis of symmetry.
Inputs
Must be non-zero. If a = 0, the equation is linear, not quadratic.
Root 1 (x₁)
3
Root 2 (x₂)
2
Root type
two distinct real roots
Discriminant (D = b²−4ac)
1.000000
Axis of symmetry (x = −b/2a)
2.500000
Vertex x-coordinate
2.500000
Vertex y-coordinate
-0.250000
Step by step
Identify coefficients
= a = 1, b = -5, c = 6
Discriminant
D = b² − 4ac = -5² − 4(1)(6) = 25 − 24
= D = 1
√D
= √1 = 1
Root 1
x₁ = (−-5 + 1) / (2 × 1)
= x₁ = 3
Root 2
x₂ = (−-5 − 1) / (2 × 1)
= x₂ = 2
Vertex
(−b/2a, f(−b/2a)) = (2.5, -0.25)
= (2.5, -0.25)
How it works
The quadratic formula solves ax² + bx + c = 0 for any real coefficients. The discriminant b² − 4ac determines the nature of the roots: positive gives two distinct real roots, zero gives one repeated root, negative gives two complex conjugate roots. The vertex is the highest or lowest point of the parabola y = ax² + bx + c.
Formulas
Quadratic formula
x = (−b ± √(b²−4ac)) / (2a)
- a
- Coefficient of x²
- b
- Coefficient of x
- c
- Constant term
- b²−4ac
- Discriminant D
Vertex
Vertex = (−b/(2a), −D/(4a))
Frequently Asked Questions
What does the discriminant tell me?
The discriminant D = b² − 4ac reveals the nature of the roots without solving: D > 0 means two real roots; D = 0 means exactly one real root (the parabola touches the x-axis at its vertex); D < 0 means no real roots — the parabola doesn't cross the x-axis and the roots are complex.
What are complex roots?
When D < 0, the roots involve √(negative number) = imaginary numbers. They appear in conjugate pairs: a + bi and a − bi. They are mathematically valid but correspond to no real x-axis crossing.
How do I verify the roots?
Substitute each root back into ax² + bx + c. The result should be 0 (or very close to 0 due to floating-point precision). You can also verify by multiplying the factored form: a(x − x₁)(x − x₂) should expand to ax² + bx + c.