How to Use the Quadratic Formula: Step-by-Step With Examples

The quadratic formula solves any equation of the form ax² + bx + c = 0. Plug the coefficients into x = (-b ± √(b² - 4ac)) / (2a) and simplify. That's it — the formula always works, even when factoring fails.

Below we walk through two full examples, break the method into five clean steps, and flag the mistakes students most often make with signs and the discriminant.

Worked Example 1: Solve 2x² + 3x − 5 = 0

Identify the coefficients directly from the equation: a = 2, b = 3, c = -5.

1. Write the formula: x = (-b ± √(b² - 4ac)) / (2a).
2. Compute the discriminant: b² - 4ac = 3² - 4(2)(-5) = 9 + 40 = 49.
3. Take the square root: √49 = 7.
4. Plug in: x = (-3 ± 7) / (2·2) = (-3 ± 7) / 4.
5. Split the ± into two solutions: x = (-3 + 7)/4 = 4/4 = 1 and x = (-3 - 7)/4 = -10/4 = -5/2.

Solutions: x = 1 and x = -5/2.

The Method in 5 Steps

1. Rewrite the equation so one side is zero. Example: x² = 6x - 9 becomes x² - 6x + 9 = 0.
2. Identify a, b, and c. For 3x² - 4x + 1 = 0, that's a = 3, b = -4, c = 1. Keep the signs attached.
3. Compute the discriminant D = b² − 4ac. For a=1, b=-4, c=1: D = 16 - 4 = 12.
4. Plug into x = (-b ± √D) / (2a). Watch the negative on -b: if b = -6, then -b = 6.
5. Simplify each branch separately. One equation with ± becomes two clean solutions.

What the discriminant tells you

• D > 0: two distinct real solutions.
• D = 0: one repeated real solution.
• D < 0: two complex solutions (involving i).

Worked Example 2: Solve x² − 6x + 4 = 0

Here a = 1, b = -6, c = 4.

1. Discriminant: b² - 4ac = (-6)² - 4(1)(4) = 36 - 16 = 20.
2. Simplify the radical: √20 = √(4·5) = 2√5.
3. Apply the formula: x = -(-6) ± 2√5) / (2·1) = (6 ± 2√5) / 2.
4. Reduce: x = 3 ± √5.

Solutions: x = 3 + √5 and x = 3 - √5. Numerically, that's about 5.236 and 0.764.

Common Mistakes

• Dropping the negative on b. The formula starts with -b, so if b = -6, you write +6, not -6.
• Forgetting to set the equation to zero first. You cannot read off a, b, c from 2x² + 3x = 5 — move the 5 over first.
• Sign errors in −4ac. If c is negative, -4ac becomes positive and increases the discriminant.
• Dividing only part of the numerator by 2a. Both -b and √D get divided — use parentheses or a fraction bar.
• Simplifying the radical too late. Reduce √20 to 2√5 before dividing, so the final fraction cancels cleanly.

You now have the formula, the discriminant check, and the two-branch split. The fastest way to lock it in is to work a few problems where a coach nudges you the moment you drop a sign — not after you've written the whole answer. Try a guided quadratic below and see the method click.