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Quadratic Formula Calculator

Type the values of a, b and c to solve ax² + bx + c = 0.

Equation

x² - 3x + 2 = 0

Roots

x₁ = 2

x₂ = 1

Steps

1. Quadratic formula: x = (−b ± √(b² − 4ac)) / 2a

2. Substitute: x = (−(-3) ± √((-3)² − 4 × 1 × 2)) / (2 × 1)

3. Discriminant: b² − 4ac = 1

4. The discriminant is positive, so there are two real roots.

Vertex(1.5, -0.25)
Axis of symmetryx = 1.5

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Solving quadratic equations

A quadratic is any equation you can write as ax² + bx + c = 0, where a is not zero. Its graph is a parabola, a smooth U shape. Solving it means finding where that curve crosses the x-axis, and those crossing points are the roots.

The quadratic formula works on every quadratic without exception, which is why it is worth memorising. Factoring is faster when it works, but it does not always work.

The quadratic formula

Read a, b and c straight off the equation once it is arranged with zero on one side. Watch the signs: in x² − 3x + 2 = 0 the value of b is −3, not 3.

The formulax = ( −b ± √(b² − 4ac) ) / 2a

The ± is why a quadratic usually has two answers. You work the whole calculation twice, once adding the square root and once subtracting it.

What the discriminant tells you

The part under the square root, b² − 4ac, is called the discriminant. Calculating it first tells you what kind of answer to expect, which is a useful check before you do the rest of the arithmetic.

  • Positive: two different real roots. The parabola crosses the x-axis twice.
  • Zero: one repeated real root. The parabola just touches the x-axis at its vertex.
  • Negative: no real roots. The parabola never reaches the x-axis, and the two answers are complex numbers.

A negative discriminant is not a mistake. Exam questions often ask you to show that an equation has no real roots, and the discriminant is the whole answer.

A worked example, step by step

Take 2x² + 5x − 3 = 0, so a = 2, b = 5 and c = −3.

1. Substitutex = ( −5 ± √(5² − 4 × 2 × −3) ) / (2 × 2)
2. Discriminant25 − (−24) = 49
3. Square root√49 = 7
4. Both branchesx = (−5 + 7) / 4 = 0.5 and x = (−5 − 7) / 4 = −3

Check your answers by substituting them back. Putting x = 0.5 into 2x² + 5x − 3 gives 0.5 + 2.5 − 3 = 0, so it holds.

When to factor instead

If a is 1 and you can spot two numbers that multiply to c and add to b, factoring is quicker. For x² − 3x + 2 = 0 those numbers are −1 and −2, giving (x − 1)(x − 2) = 0 and roots of 1 and 2.

A perfect square discriminant is the sign that a quadratic factors neatly. In the example above it was 1, and in the worked example it was 49. When the discriminant is not a perfect square, the roots are irrational and factoring will not help.

The vertex and the axis of symmetry

Every parabola is symmetrical about a vertical line through its lowest or highest point, and that point is the vertex. Its x-coordinate sits exactly halfway between the two roots, at −b divided by 2a.

Put that value back into the equation to get the y-coordinate. When a is positive the parabola opens upward and the vertex is the minimum. When a is negative it opens downward and the vertex is the maximum, which is what word problems about maximum height or maximum profit are really asking for.

Completing the square and vertex form

Vertex form is a(x − h)² + k, where (h, k) is the vertex. It is useful because you can read the vertex straight off it, and sketching the parabola becomes easy.

Completing the square is how you convert to it, and it is also where the quadratic formula comes from. If you complete the square on the general equation ax² + bx + c = 0 and solve for x, the formula falls out. Do that once and the formula stops being something you memorise blindly.

Mistakes that cost marks

  • Forgetting to rearrange first. The formula needs zero on one side, so 2x² + 5x = 3 must become 2x² + 5x − 3 = 0.
  • Losing a sign on b. In x² − 3x + 2, b is −3, and −b is therefore +3.
  • Squaring a negative b incorrectly. (−3)² is 9, not −9.
  • Dividing by 2 instead of 2a. When a = 2 the denominator is 4.
  • Dividing only part of the numerator. The whole of −b ± √(b² − 4ac) is divided by 2a.
  • Stopping at one root when the discriminant is positive.

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