About this calculator
Enter the three numbers a, b and c from your quadratic equation ax² + bx + c = 0 and this solver finds the solutions, with the working laid out step by step so you can follow it or check your own.
It handles every case: two different real solutions, one repeated solution, and equations with no real solutions, which have two complex ones. It also gives the vertex of the parabola.
Worked examples
Real numbers, worked out by the same calculator. Press “Use these numbers” to try one above.
x² − 5x + 6 = 0
- Solutions
- x = 2 or x = 3
- Discriminant (b² − 4ac)
- 1
- Type of solutions
- Two different real solutions
- Vertex of the parabola
- (2.5, -0.25)
x² − 5x + 6 = 0 has the solutions x = 2 or x = 3.
Show the working
- Write it as ax² + bx + c = 0, so a = 1, b = -5 and c = 6. (Your equation: x² − 5x + 6 = 0.)
- Find the discriminant: b² − 4ac = (-5)² − 4 × 1 × 6 = 1.
- The discriminant is positive, so there are two real solutions.
- x = (−b ± √discriminant) ÷ 2a = (5 ± √1) ÷ 2.
- So x = 2 or x = 3.
2x² + 3x − 2 = 0
- Solutions
- x = -2 or x = 0.5
- Discriminant (b² − 4ac)
- 25
- Type of solutions
- Two different real solutions
- Vertex of the parabola
- (-0.75, -3.125)
2x² + 3x − 2 = 0 has the solutions x = -2 or x = 0.5.
Show the working
- Write it as ax² + bx + c = 0, so a = 2, b = 3 and c = -2. (Your equation: 2x² + 3x − 2 = 0.)
- Find the discriminant: b² − 4ac = 3² − 4 × 2 × -2 = 25.
- The discriminant is positive, so there are two real solutions.
- x = (−b ± √discriminant) ÷ 2a = (-3 ± √25) ÷ 4.
- So x = -2 or x = 0.5.
x² − 4x + 4 = 0 (one repeated solution)
- Solutions
- x = 2
- Discriminant (b² − 4ac)
- 0
- Type of solutions
- One repeated real solution
- Vertex of the parabola
- (2, 0)
x² − 4x + 4 = 0 has the solutions x = 2.
Show the working
- Write it as ax² + bx + c = 0, so a = 1, b = -4 and c = 4. (Your equation: x² − 4x + 4 = 0.)
- Find the discriminant: b² − 4ac = (-4)² − 4 × 1 × 4 = 0.
- The discriminant is zero, so there is one repeated solution.
- x = −b ÷ 2a = 4 ÷ 2 = 2.
x² + 2x + 5 = 0 (no real solutions)
- Solutions
- x = -1 + 2i or x = -1 − 2i
- Discriminant (b² − 4ac)
- -16
- Type of solutions
- Two complex solutions (no real solutions)
- Vertex of the parabola
- (-1, 4)
x² + 2x + 5 = 0 has the solutions x = -1 + 2i or x = -1 − 2i.
Show the working
- Write it as ax² + bx + c = 0, so a = 1, b = 2 and c = 5. (Your equation: x² + 2x + 5 = 0.)
- Find the discriminant: b² − 4ac = 2² − 4 × 1 × 5 = -16.
- The discriminant is negative, so there are no real solutions, only complex ones.
- x = (−b ± √discriminant) ÷ 2a = -1 ± 2i, where i is the square root of −1.
The quadratic formula
For any equation ax² + bx + c = 0, the solutions are
x = (−b ± √(b² − 4ac)) ÷ 2a
The part under the square root, b² − 4ac, is called the discriminant. It tells you what kind of answer to expect before you finish the calculation.
| Discriminant | Solutions |
|---|---|
| Positive | Two different real solutions (the parabola crosses the x-axis twice) |
| Zero | One repeated real solution (the parabola just touches the x-axis) |
| Negative | No real solutions; two complex solutions (the parabola doesn't reach the x-axis) |
Checking your answer
Two quick checks catch most mistakes. First, substitute a solution back into the equation: for x² − 5x + 6 = 0 and x = 2, you get 4 − 10 + 6 = 0. Second, use the fact that the two solutions add up to −b ÷ a and multiply to c ÷ a. Here, 2 + 3 = 5 (that's −(−5) ÷ 1) and 2 × 3 = 6.
Common mistakes
- Sign errors with b. If b is −5, then −b is +5. Squaring a negative b gives a positive number, so b² is (−5)² = 25, not −25.
- Not rearranging first. The equation must equal zero. To solve x² + 2x = 8, move the 8 across to get x² + 2x − 8 = 0, so c = −8.
- Dividing by 2 instead of 2a. The whole top line is divided by 2a, not just by 2.
- a = 0. Then there's no x² term and it isn't a quadratic. It's a straight-line equation, bx + c = 0, with the single solution x = −c ÷ b.
What the vertex means
The graph of a quadratic is a parabola, and its vertex is the turning point: the lowest point if a is positive and the highest if a is negative. The vertex is at x = −b ÷ 2a. Putting that x back into the equation gives its height. In x² − 5x + 6 the vertex is at (2.5, −0.25).
Frequently asked questions
How do I solve a quadratic equation?
Write it as ax² + bx + c = 0, work out the discriminant b² − 4ac, and put the numbers into x = (−b ± √discriminant) ÷ 2a. This solver shows each of those steps.
What does the discriminant tell me?
Whether there are two real solutions (positive), one repeated solution (zero) or no real solutions (negative). You can find it before doing the rest of the calculation.
What are complex solutions?
When the discriminant is negative, the square root involves the imaginary number i (the square root of −1). The solutions come as a pair, such as −1 + 2i and −1 − 2i. They don't appear where the graph crosses the x-axis, because it doesn't cross.
What if there is no c or no b?
Leave that box empty and it counts as zero. For x² − 9 = 0, use a = 1, b empty and c = −9, which gives x = 3 or x = −3.
Why can't a be zero?
With a = 0 the x² term disappears, and the equation is no longer quadratic. It becomes bx + c = 0, which has only one solution.
Formulas tested against hand-worked answers. Last reviewed 29 September 2026. These calculators do arithmetic only; they are not financial, tax or legal advice.