Matematica

Quadratic Equations: Solved Exercises From Easy to Hard

How to solve a quadratic equation with the quadratic formula, step by step, plus four solved exercises of increasing difficulty with verified answers.

Recommended for: Grade 9 · Grade 10

A quadratic equation in the form ax² + bx + c = 0 (with a ≠ 0) is solved with the quadratic formula: x = (−b ± √Δ) / 2a, where Δ = b² − 4ac is the discriminant. The sign of Δ tells you immediately how many solutions to expect.

The discriminant decides everything

Before computing the roots, look at the discriminant Δ = b² − 4ac:

  • if Δ > 0 there are two distinct real roots;
  • if Δ = 0 there is one real solution (a repeated root), x = −b / 2a;
  • if Δ < 0 there are no real solutions.

The three steps

The method never changes. First rewrite the equation in standard form ax² + bx + c = 0 and identify a, b, c with their signs. Then compute Δ and take its square root. Finally apply the formula, keeping the two ± cases separate to get both roots.

The exercises below apply the same scheme with increasing difficulty: they start with simple coefficients (a = 1) and end with an equation where a ≠ 1 that needs a little care. Every answer is verified by substituting the roots back into the original equation.

Solved exercises

1. Solve: x² − 5x + 6 = 0 base

Show solution
  1. Coefficients: a = 1, b = −5, c = 6.
  2. Discriminant: Δ = b² − 4ac = 25 − 24 = 1.
  3. Δ > 0, so two distinct real roots. √Δ = 1.
  4. x = (−b ± √Δ) / 2a = (5 ± 1) / 2.
  5. x₁ = 6/2 = 3 ; x₂ = 4/2 = 2.

Answer: x = 2 ; x = 3

2. Solve: x² − 7x + 10 = 0 base

Show solution
  1. a = 1, b = −7, c = 10.
  2. Δ = 49 − 40 = 9 ; √Δ = 3.
  3. x = (7 ± 3) / 2.
  4. x₁ = 10/2 = 5 ; x₂ = 4/2 = 2.

Answer: x = 2 ; x = 5

3. Solve: x² + 2x − 8 = 0 intermedio

Show solution
  1. a = 1, b = 2, c = −8.
  2. Δ = 4 + 32 = 36 ; √Δ = 6.
  3. x = (−2 ± 6) / 2.
  4. x₁ = 4/2 = 2 ; x₂ = −8/2 = −4.

Answer: x = 2 ; x = −4

4. Solve: 2x² − 4x − 6 = 0 avanzato

Show solution
  1. a = 2, b = −4, c = −6.
  2. Δ = 16 + 48 = 64 ; √Δ = 8.
  3. x = (4 ± 8) / 4.
  4. x₁ = 12/4 = 3 ; x₂ = −4/4 = −1.

Answer: x = −1 ; x = 3

FAQ

When does a quadratic equation have no real solutions?

When the discriminant Δ = b² − 4ac is negative: the solutions are complex conjugates, not real numbers.

What happens when the discriminant equals zero?

If Δ = 0 the equation has a single real solution (a repeated root), equal to x = −b / 2a.

Do I always need the quadratic formula?

No. If the constant or the linear term is missing, factoring or a direct square root is faster; the quadratic formula is the general method that always works.