Quadratic Equation Calculator
Enter the coefficients a, b and c and this calculator solves ax² + bx + c = 0 — real or complex roots, the discriminant, and the parabola's vertex, axis and y-intercept, with every step shown.
Outside the supported range
Two distinct real solutions
Formula: x = (−b ± √(b² − 4ac)) ÷ 2a
The parabola opens upward, so the vertex is a minimum.
| Step | Value |
|---|---|
| Equation | x² − 5x + 6 = 0 |
| a | 1 |
| b | −5 |
| c | 6 |
| Discriminant formula | b² − 4ac |
| Discriminant | (−5)² − 4 × 1 × 6 = 1 |
| Quadratic formula | x = (−b ± √(b² − 4ac)) ÷ 2a |
| First solution | (5 + √1) ÷ 2 = 3 |
| Second solution | (5 − √1) ÷ 2 = 2 |
How to use this calculator
- Enter the three coefficients from ax² + bx + c = 0. Use
afor the number in front of x²,bfor the number in front of x, andcfor the constant. Negative and decimal values are fine;acannot be zero. - Check the tidied equation shown below the inputs matches what you meant.
- Read the result: the solutions are the large figures, with the discriminant and the parabola's vertex, axis of symmetry and y-intercept beside them, and the full working in the calculation table.
The calculation runs in the page as you type — nothing is sent to a server. This is a mathematical tool, not a substitute for showing your own method in an exam.
What is standard quadratic form?
A quadratic equation is any equation that can be written with a single squared term and no higher power:
- a — the coefficient of x², which must not be zero
- b — the coefficient of x, which may be zero
- c — the constant term, which may be zero
If a were zero the x² term would vanish and the equation would be linear, so a non-zeroa is what makes an equation quadratic.
How is a quadratic equation solved?
Every quadratic in standard form is solved by the quadratic formula:
- The ± sign gives the two solutions — one using +, one using −
- b² − 4ac is the discriminant, Δ
- When Δ = 0 the formula simplifies to x = −b ÷ 2a
The discriminant decides how many real solutions there are before you finish solving:
- Δ > 0 — two distinct real solutions
- Δ = 0 — one repeated real solution
- Δ < 0 — two complex solutions (a conjugate pair)
What the discriminant tells you
| Discriminant | Solutions | Example |
|---|---|---|
| Δ > 0 (positive) | Two distinct real solutions | x² − 5x + 6 = 0 → x = 3, x = 2 |
| Δ = 0 (zero) | One repeated real solution | x² − 4x + 4 = 0 → x = 2 |
| Δ < 0 (negative) | Two complex solutions | x² + 2x + 5 = 0 → x = −1 ± 2i |
Worked example: two real solutions
Solve x² − 5x + 6 = 0, where a = 1, b = −5 and c = 6.
| Step | Value |
|---|---|
| Equation | x² − 5x + 6 = 0 |
| a | 1 |
| b | −5 |
| c | 6 |
| Discriminant formula | b² − 4ac |
| Discriminant | (−5)² − 4 × 1 × 6 = 1 |
| Quadratic formula | x = (−b ± √(b² − 4ac)) ÷ 2a |
| First solution | (5 + √1) ÷ 2 = 3 |
| Second solution | (5 − √1) ÷ 2 = 2 |
The discriminant is positive, so there are two real solutions: x = 3 and x = 2. The parabola opens upward, its axis of symmetry is x = 2.5, its vertex is (2.5, −0.25) and its y-intercept is (0, 6).
Worked example: two complex solutions
Solve x² + 2x + 5 = 0, where a = 1, b = 2 and c = 5.
| Step | Value |
|---|---|
| Equation | x² + 2x + 5 = 0 |
| a | 1 |
| b | 2 |
| c | 5 |
| Discriminant | 2² − 4 × 1 × 5 = −16 |
| Result type | Negative discriminant: complex solutions |
| Quadratic formula | x = (−b ± √(b² − 4ac)) ÷ 2a |
| Real part | −2 ÷ 2 = −1 |
| Imaginary part | √16 ÷ 2 = 2 |
| Solutions | −1 ± 2i |
The discriminant is −16, which is negative, so there are no real solutions. The solutions are the complex conjugate pair −1 + 2i and −1 − 2i. The parabola still has a vertex at (−1, 4), an axis of symmetry of x = −1 and a y-intercept of (0, 5) — it simply never crosses the x-axis.
How to find the vertex and axis of symmetry
The graph of y = ax² + bx + c is a parabola, symmetric about a vertical line through its turning point. That line is the axis of symmetry and the turning point is the vertex.
- Vertex x-coordinate: h = −b ÷ 2a
- Vertex y-coordinate: k = ah² + bh + c
- Y-intercept: (0, c)
- a > 0 opens upward (vertex is a minimum); a < 0 opens downward (vertex is a maximum)
Results are shown to at most six decimal places with trailing zeroes removed, so an exact solution reads 3 and an irrational one reads 1.414214. Values within rounding distance of zero are shown as 0, never −0.