How to solve quadratic equations

A quadratic equation is any equation that can be written in the formax² + bx + c = 0, where a, b, and c are numbers and a is never 0 (if a were 0, the x² term would vanish and it would just be a linear equation). Examples includex² - 5x + 6 = 0, 2x² + 3x - 2 = 0, and x² - 9 = 0. Unlike a linear equation, which has exactly one solution, a quadratic equation typically has two.

There are three standard ways to solve one: factoring, the quadratic formula, and completing the square. This guide focuses on the first two, since between them they cover every quadratic equation you'll meet in a typical algebra course, and shows how to read the discriminant to know what kind of answer to expect before you even finish solving.

Method 1: factoring

Factoring rewrites ax² + bx + c as a product of two binomials, then uses the zero product property: if two things multiply to zero, at least one of them has to be zero. This method is fastest when a, b, and c are small whole numbers that factor cleanly.

Worked example 1: a = 1

Solve x² - 5x + 6 = 0.

1. Look for two numbers that multiply to 6 (the c term) and add to -5 (the b term). -2 and -3 work: (-2) × (-3) = 6 and (-2) + (-3) = -5.
2. Write the factored form: (x - 2)(x - 3) = 0.
3. Apply the zero product property: either x - 2 = 0 or x - 3 = 0, giving x = 2 or x = 3.
4. Check both: (2)² - 5(2) + 6 = 4 - 10 + 6 = 0, and(3)² - 5(3) + 6 = 9 - 15 + 6 = 0. Both check out.

Worked example 2: a is not 1

Solve 2x² + 3x - 2 = 0.

1. Multiply a and c: 2 × (-2) = -4. Look for two numbers that multiply to -4 and add to 3 (the b term). 4 and -1 work.
2. Split the middle term using those two numbers: 2x² + 4x - x - 2 = 0.
3. Group and factor each pair: 2x(x + 2) - 1(x + 2) = 0, which becomes(2x - 1)(x + 2) = 0.
4. Set each factor to zero: 2x - 1 = 0 gives x = 1/2, andx + 2 = 0 gives x = -2.
5. Check: 2(1/2)² + 3(1/2) - 2 = 0.5 + 1.5 - 2 = 0, and2(-2)² + 3(-2) - 2 = 8 - 6 - 2 = 0. Both correct.

Worked example 3: a difference of squares

Solve x² - 9 = 0.

1. Recognize this as a difference of squares, x² - 3², which always factors as(x - 3)(x + 3).
2. Set each factor to zero: x = 3 or x = -3.
3. Check: 3² - 9 = 0 and (-3)² - 9 = 0. Both correct.

You can also solve x² - 9 = 0 even faster by adding 9 to both sides (x² = 9) and taking the square root of both sides, remembering both the positive and negative root: x = ±3. This shortcut works any time there's no middle (b) term.

Method 2: the quadratic formula

Not every quadratic factors neatly with whole numbers. The quadratic formula solvesax² + bx + c = 0 for any a, b, and c:

x = (-b ± √(b² - 4ac)) / 2a

The ± symbol means you calculate the expression twice, once adding the square root and once subtracting it, which is where the two solutions of a quadratic usually come from.

Worked example: solving with the formula

Solve x² + 2x - 8 = 0 using the formula.

1. Identify a = 1, b = 2, c = -8.
2. Substitute into the formula: x = (-2 ± √(2² - 4(1)(-8))) / 2(1).
3. Simplify inside the square root: 2² - 4(1)(-8) = 4 + 32 = 36, sox = (-2 ± √36) / 2.
4. Since √36 = 6: x = (-2 + 6) / 2 = 2, orx = (-2 - 6) / 2 = -4.
5. Check: (2)² + 2(2) - 8 = 4 + 4 - 8 = 0, and(-4)² + 2(-4) - 8 = 16 - 8 - 8 = 0. Both correct.

Notice this equation would also have factored quickly ((x + 4)(x - 2) = 0), so the formula and factoring agree, as they always will. The formula is most valuable when factoring isn't obvious or doesn't exist in whole numbers.

Worked example: irrational solutions

Solve x² - 4x - 1 = 0.

1. Identify a = 1, b = -4, c = -1.
2. Substitute: x = (4 ± √((-4)² - 4(1)(-1))) / 2(1) = (4 ± √(16 + 4)) / 2 = (4 ± √20) / 2.
3. Simplify the radical: √20 = 2√5, so x = (4 ± 2√5) / 2 = 2 ± √5.
4. This equation doesn't factor with whole numbers, which is exactly the situation the quadratic formula is built for. The two exact answers are x = 2 + √5 andx = 2 - √5, or approximately 4.24 and -0.24.

Reading the discriminant

The expression under the square root in the quadratic formula, b² - 4ac, is called the discriminant. You can compute it before finishing the rest of the formula to know what kind of answer to expect:

For x² + 2x - 8 = 0 above, the discriminant was 36, a positive perfect square, which is why the answers came out as clean whole numbers. For x² - 4x - 1 = 0, the discriminant was 20, positive but not a perfect square, which is why the answers involved a radical instead.

Worked example: a negative discriminant

Solve x² + x + 1 = 0.

1. Identify a = 1, b = 1, c = 1.
2. Discriminant: 1² - 4(1)(1) = 1 - 4 = -3.
3. Since the discriminant is negative, this equation has no real solutions. Its graph, a parabola opening upward, never crosses the x-axis. If your course covers complex numbers, the two solutions are x = (-1 ± i√3) / 2; otherwise, "no real solution" is the complete answer.

Common mistakes

• Forgetting to set the equation to zero first. Factoring and the quadratic formula both require the form ax² + bx + c = 0. An equation likex² + 3x = 10 needs to become x² + 3x - 10 = 0 before you factor or apply the formula.
• Dropping the negative sign on b. In the formula, it's -b, not b. If b is already negative, like in x² - 4x - 1 where b = -4, then -b becomes positive 4.
• Only keeping one of the two solutions. Both the plus and minus versions of the ± are usually valid answers. Dropping one loses half the solution set.
• Simplifying the square root incorrectly. √20 simplifies to2√5, not 10√2 or left as an unsimplified radical, when an exact answer is expected.
• Trying to factor an equation that doesn't factor with whole numbers. If a few honest attempts at factoring don't produce whole-number pairs, switch to the quadratic formula instead of guessing indefinitely.

Where quadratics show up outside of class

Quadratic equations model anything where a quantity changes according to the square of another quantity: the height of a ball thrown into the air over time, the area of a rectangular space given a fixed perimeter, and the relationship between price and revenue when raising a price reduces the number of buyers. In each case, setting the equation equal to a target value and solving for the unknown is exactly the process worked through above.

Practice it yourself

Try the math solver with your own quadratic equation to see every step, including the discriminant, laid out automatically. For repeated practice, head toquadratic equation practice, or reviewlinear equations first if the balance principle behind moving terms across the equals sign still feels shaky.