Solving Quadratic Equations

A quadratic equation has the form

a·x² + b·x + c = 0, where a ≠ 0

Its graph is a parabola, and solving the equation means finding where that parabola crosses the x-axis.

Method 1: Factoring

Solve x² − 5x + 6 = 0.

Look for two numbers that multiply to 6 and add to −5: those are −2 and −3.

(x − 2)(x − 3) = 0, so x = 2 or x = 3

Factoring is fastest when the roots are whole numbers, but it does not always work neatly.

Method 2: The quadratic formula

Every quadratic can be solved with:

x = (−b ± √(b² − 4ac)) ÷ 2a

Solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2.

1. b² − 4ac = 9 + 16 = 25
2. √25 = 5
3. x = (−3 + 5) ÷ 4 = 0.5 or x = (−3 − 5) ÷ 4 = −2

So x = 0.5 or x = −2.

What the discriminant tells you

The expression D = b² − 4ac is called the discriminant, and it predicts the type of roots before you solve:

Discriminant	Roots
D > 0	Two distinct real roots
D = 0	One repeated real root
D < 0	Two complex conjugate roots

Example: x² + x + 1 = 0 has D = 1 − 4 = −3 < 0, so it has no real solutions — its parabola never touches the x-axis.

Method 3: Completing the square

Rewrite the equation so one side is a perfect square. For x² + 6x + 5 = 0:

1. Move the constant: x² + 6x = −5
2. Add (6/2)² = 9 to both sides: x² + 6x + 9 = 4
3. Factor: (x + 3)² = 4
4. So x + 3 = ±2, giving x = −1 or x = −5

This method is how the quadratic formula is derived, and it is the standard form used when sketching parabolas.

Common mistakes

• Forgetting that a cannot be zero — otherwise the equation is linear, not quadratic.
• Sign errors when substituting negative coefficients into the formula.
• Stopping at the discriminant: D tells you about the roots, but you still need to compute them.