Deriving the Quadratic Formula- Step-by-Step Proof

Why Derive the Quadratic Formula Instead of Just Memorizing It

Most students memorize ax² + bx + c = 0 and the formula that follows. Fewer understand why it works. That's a problem because the derivation uses one technique—completing the square—that shows up everywhere in algebra.

You don't need to derive this formula on a test. But if you understand the logic, quadratic equations stop being a mystery. You see what you're actually doing when you solve them.

The Setup: Standard Form

Every quadratic equation starts here:

ax² + bx + c = 0

Where a, b, and c are constants, and a ≠ 0. If a = 0, you don't have a quadratic—you have a linear equation.

The goal is to isolate x. The formula you get at the end gives you x in terms of a, b, and c.

The Derivation: Completing the Square

Here's every step. No shortcuts skipped.

Step 1: Move the constant term

Start with:

ax² + bx + c = 0

Subtract c from both sides:

ax² + bx = -c

Step 2: Factor out the leading coefficient

Factor a from the left side:

a(x² + (b/a)x) = -c

Step 3: Prepare to complete the square

Take the coefficient of x, which is b/a. Divide it by 2, then square it:

(b/2a)² = b²/4a²

add this value inside the parentheses. But you can't just add it—you have to subtract it too, so the expression stays equal:

a(x² + (b/a)x + b²/4a² - b²/4a²) = -c

Step 4: Rewrite as a perfect square

The first three terms inside parentheses form a perfect square trinomial:

x² + (b/a)x + b²/4a² = (x + b/2a)²

So now you have:

a((x + b/2a)² - b²/4a²) = -c

Step 5: Distribute and isolate the squared term

Multiply a through:

a(x + b/2a)² - b²/4a = -c

Add b²/4a to both sides:

a(x + b/2a)² = -c + b²/4a

Step 6: Combine the right side

Get a common denominator to add the terms on the right:

-c = -4ac/4a

So:

-c + b²/4a = (b² - 4ac)/4a

Your equation is now:

a(x + b/2a)² = (b² - 4ac)/4a

Step 7: Divide both sides by a

(x + b/2a)² = (b² - 4ac)/(4a²)

Step 8: Take the square root of both sides

Remember: when you take the square root, you get ±:

x + b/2a = ±√[(b² - 4ac)/(4a²)]

Simplify the denominator:

√(4a²) = 2|a|

Since we're working with the ± anyway, we can write this as:

x + b/2a = ±√(b² - 4ac) / 2a

Step 9: Solve for x

Subtract b/2a from both sides:

x = -b/2a ± √(b² - 4ac) / 2a

Combine over a single denominator:

x = [-b ± √(b² - 4ac)] / 2a

The Quadratic Formula

That's it. Here's the result:

x = (-b ± √(b² - 4ac)) / 2a

The expression under the square root—b² - 4ac—is called the discriminant. It tells you how many real solutions you'll get.

Discriminant Breakdown

How to Use the Formula

Let's work through an example:

2x² + 5x - 3 = 0

Identify your values:

• a = 2
• b = 5
• c = -3

Plug into the formula:

x = [-5 ± √(25 - 4(2)(-3))] / 2(2)

Simplify under the square root:

25 - 4(2)(-3) = 25 + 24 = 49

√49 = 7

Now solve both versions:

x = (-5 + 7) / 4 = 2/4 = 0.5

x = (-5 - 7) / 4 = -12/4 = -3

Your solutions are x = 0.5 and x = -3.

Where Students Go Wrong

• Forgetting the ± sign. The ± gives you two solutions. Skipping it means you only get one.
• Screwing up negative signs when substituting. Double-check that b = -3 means you plug in -3, not 3.
• Simplifying the square root incorrectly. √(49) = 7, not ±7. The ± is already in the formula.
• Dividing by 2a incorrectly. Make sure the entire numerator sits over 2a, not just part of it.

When to Use the Formula vs. Other Methods

The formula always works. Factoring only works when the roots are rational numbers. If you can't factor it quickly, just use the formula.