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Quadratic Equation Solver (Quadratic Formula)

The quadratic formula is x = (−b ± √(b² − 4ac)) ÷ 2a. For x² − 5x + 6 = 0, the discriminant is 25 − 24 = 1, so x = (5 ± 1) ÷ 2, giving x = 3 or x = 2.

ax² + bx + c = 0

x² − 5x + 6 = 0

x₁

2

x₂

3

Discriminant (b² − 4ac)

1

Vertex

(2.5, -0.25)

Show the math

  1. 1

    Read off a, b and c

    x² − 5x + 6 = 0 → a = 1, b = -5, c = 6

    Compare with the standard form ax² + bx + c = 0.

Read the steps as text
  1. Read off a, b and c. x² − 5x + 6 = 0 → a = 1, b = -5, c = 6 Compare with the standard form ax² + bx + c = 0.
  2. Discriminant = b² − 4ac. (-5)² − 4 × 1 × 6 = 25 − 24 = 1 Positive, so there are two different real roots.
  3. √Discriminant. √1 = 1
  4. x = (−b ± √Δ) ÷ 2a. (5 + 1) ÷ 2 = 3 · (5 − 1) ÷ 2 = 2 The ± gives one root with + and one with −.
  5. Vertex of the parabola. x = −b ÷ 2a = 2.5, y = -0.25 The turning point of the curve. It's a minimum, since a > 0 and the parabola opens upward.

Using the quadratic formula

Any equation of the form ax² + bx + c = 0, where a isn't zero, can be solved with the quadratic formula: x = (−b ± √(b² − 4ac)) ÷ 2a. The ± means there are usually two answers, one using + and one using −.

The calculator reads off a, b and c, works out the discriminant b² − 4ac, takes its square root and substitutes everything into the formula. Make sure the equation equals zero first: 2x² = 3x + 2 has to be rearranged to 2x² − 3x − 2 = 0, so a = 2, b = −3 and c = −2.

What the discriminant tells you

The discriminant, b² − 4ac, decides what kind of roots you get. If it's positive, there are two different real roots, where the parabola crosses the x-axis twice. If it's zero, there's one repeated root, where the parabola just touches the axis. If it's negative, there are no real roots; the two roots are complex numbers of the form p ± q𝑖, where 𝑖 = √−1.

The vertex, at x = −b ÷ 2a, is the parabola's turning point: its lowest point when a is positive and its highest when a is negative. For x² + 2x + 5 = 0 the discriminant is 4 − 20 = −16, so the roots are −1 ± 2𝑖 and the vertex is at (−1, 4), above the x-axis.

Frequently asked questions

What is the quadratic formula?
x = (−b ± √(b² − 4ac)) ÷ 2a, for any equation ax² + bx + c = 0 with a ≠ 0.
What if the discriminant is negative?
There are no real solutions. The equation has two complex roots, −b ÷ 2a ± (√(4ac − b²) ÷ 2a)𝑖, which the calculator shows.
Can I solve by factoring instead?
When the roots are nice whole numbers, yes: x² − 5x + 6 = (x − 2)(x − 3). The quadratic formula always works, including when the roots are decimals or complex numbers that are hard to spot by factoring.
Why can't a be zero?
Without the x² term the equation is linear, bx + c = 0, with the single solution x = −c ÷ b.

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