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Quadratic Equations

A quadratic equation has the form ax² + bx + c = 0. It can have zero, one, or two real solutions. The parabola it describes is one of the most common curves in nature and physics — from the path of a thrown ball to satellite dish shapes.

In this lesson

  1. What makes an equation quadratic
  2. Method 1: Solving by factoring
  3. Method 2: The quadratic formula
  4. Method 3: Completing the square
  5. The discriminant: predicting the number of solutions

1 What Makes an Equation Quadratic

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a ≠ 0. The highest power of x is 2. Examples: x² − 5x + 6 = 0, 2x² + 3x = 0, x² = 16.

Quadratics can have up to two solutions (also called roots or zeros) because a parabola can cross the x-axis at 0, 1, or 2 points. Understanding all three methods for solving them gives you flexibility for any situation.

2 Method 1: Solving by Factoring

If you can factor the quadratic, set each factor equal to zero and solve. This works because if A × B = 0, then A = 0 or B = 0 (the zero product property).

Solving by Factoring
Solve: x² + 5x + 6 = 0
1Factor the left side: (x + 2)(x + 3) = 0
2Set each factor to zero: x + 2 = 0 → x = −2
3And: x + 3 = 0 → x = −3
4Check x=−2: (−2)²+5(−2)+6 = 4−10+6 = 0 ✓
Answer: x = −2 or x = −3

3 Method 2: The Quadratic Formula

The quadratic formula always works, even when factoring is difficult or impossible: x = (−b ± √(b² − 4ac)) / 2a

Using the Quadratic Formula
Solve: 2x² − 4x − 6 = 0
1Identify a=2, b=−4, c=−6
2b² − 4ac = (−4)² − 4(2)(−6) = 16 + 48 = 64
3x = (4 ± √64) / 4 = (4 ± 8) / 4
4x = (4+8)/4 = 3 OR x = (4−8)/4 = −1
Answer: x = 3 or x = −1
When to use the formula

Use the quadratic formula when: the equation doesn't factor nicely, the coefficients are large or decimal, or you need an exact answer in terms of radicals. It always produces the correct answer.

4 Method 3: Completing the Square

Completing the square rewrites the quadratic in the form (x + h)² = k, then solves by taking the square root. This method is important for deriving the quadratic formula and for understanding vertex form of parabolas.

Completing the Square
Solve: x² + 6x + 5 = 0
1Move the constant: x² + 6x = −5
2Add (6/2)² = 9 to both sides: x² + 6x + 9 = −5 + 9 = 4
3Factor the perfect square: (x + 3)² = 4
4Take the square root: x + 3 = ±2
5x = −3 + 2 = −1 OR x = −3 − 2 = −5
Answer: x = −1 or x = −5

5 The Discriminant: Predicting Solutions

The expression under the square root in the quadratic formula — b² − 4ac — is called the discriminant. It tells you how many real solutions exist before you solve:

Discriminant outcomes
2b² − 4ac > 0: Two distinct real solutions. The parabola crosses the x-axis at two points.
1b² − 4ac = 0: Exactly one real solution (a repeated root). The parabola touches the x-axis at exactly one point.
0b² − 4ac < 0: No real solutions. The parabola doesn't cross the x-axis (solutions are complex numbers).

Try the Quadratic Formula Calculator

Enter a, b, and c — get both roots instantly with the discriminant shown.

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Practice Problems

Solve: x² − 7x + 12 = 0 by factoring
Factor: (x − 3)(x − 4) = 0. Solutions: x = 3 or x = 4
How many solutions does x² + 4x + 5 = 0 have?
Discriminant = 4² − 4(1)(5) = 16 − 20 = −4. Since discriminant < 0, there are no real solutions.
Solve: 3x² − 7x + 2 = 0 using the quadratic formula
a=3, b=−7, c=2. Discriminant = 49−24 = 25. x = (7±5)/6. x = 12/6 = 2 or x = 2/6 = 1/3.