Completing the square is the method that saves you when a quadratic equation refuses to factor neatly. It turns ax^2 + bx + c = 0 into a perfect squared binomial like (x + d)^2 = k, which you can then solve with square roots. It also reveals the vertex of a parabola, making it useful far beyond equation-solving. Here is the full procedure, worked out step by step.
What Does Completing the Square Mean?
The goal is to rearrange the equation so one side is a perfect square trinomial — an expression that collapses into two identical factors:
- x² + 6x + 9 = (x + 3)²
- x² − 10x + 25 = (x − 5)²
The engine behind the whole method is one small formula. To turn x² + bx into a perfect square, add:
(b/2)^2
Half the middle coefficient, squared. That is all you need to memorize.
If you need a refresher on splitting terms first, explore our tutorial on how to factor polynomials with examples. For instant verification of roots and constants, use the CalcSolver computation engine.
The 5-Step Framework
- Make the leading coefficient 1. If a ≠ 1, divide every term in the equation by a.
- Isolate the variable terms. Keep the x-terms on the left; move the constant to the right.
- Compute (b/2)². Halve coefficient b, square it, and add the result to both sides.
- Factor the left side. Write it as a squared binomial: (x + b/2)².
- Take square roots. Apply the ± sign and solve for x.
Example 1: x² + 6x − 7 = 0
Step 1: Move the constant
Add 7 to both sides:
x² + 6x = 7
Step 2: Compute (b/2)²
Here b = 6. Half of 6 is 3, and 3² = 9.
Step 3: Add 9 to both sides
x² + 6x + 9 = 7 + 9
x² + 6x + 9 = 16
Step 4: Collapse into a squared binomial
(x + 3)² = 16
Step 5: Solve by square roots
x + 3 = ±4
- Addition branch: x = −3 + 4 = 1
- Subtraction branch: x = −3 − 4 = −7
Final answer: x = 1 or x = −7. (Check: 1 + 6 − 7 = 0 ✓; 49 − 42 − 7 = 0 ✓)
Example 2: x² + 3x − 1 = 0 (Fractions and Messy Roots)
Not every constant produces a clean square root. That is fine — the method works identically.
Step 1: Move the constant
x² + 3x = 1
Step 2: Calculate (b/2)²
Here b = 3. Half of 3 is 3/2, and (3/2)² = 9/4.
Step 3: Add 9/4 to both sides
x² + 3x + (9/4) = 1 + (9/4)
Convert 1 to fourths: 1 = 4/4, so 4/4 + 9/4 = 13/4.
x² + 3x + (9/4) = 13/4
Step 4: Factor the left side
(x + 3/2)² = 13/4
Step 5: Extract square roots
x + 3/2 = ±√(13/4) = ±(√13)/2
x = (−3 ± √13)/2
Working with fourths and halves is routine here, so solid fraction sense helps. See our guide on converting 1/4 as a percent and decimal and our reference on what is equivalent to 3/4 if you want to strengthen that foundation.
Example 3: Leading Coefficient ≠ 1 — 2x² − 8x + 6 = 0
Step 1: Divide everything by 2 first.
x² − 4x + 3 = 0
Step 2: Move the constant.
x² − 4x = −3
Step 3: Compute (b/2)². Here b = −4. Half of −4 is −2, and (−2)² = 4.
Step 4: Add 4 to both sides and factor.
x² − 4x + 4 = −3 + 4 → x² − 4x + 4 = 1 → (x − 2)² = 1
Step 5: Solve.
x − 2 = ±1
- x = 2 + 1 = 3
- x = 2 − 1 = 1
Final answer: x = 3 or x = 1. (Check: 2(9) − 8(3) + 6 = 0 ✓; 2 − 8 + 6 = 0 ✓)
If the middle term is negative, the binomial stays negative: x² − 8x completes to (x − 4)², not (x + 4)². Keep the sign of b.
Critical Mistakes to Avoid
- Failing to balance both sides: add (b/2)² to the left and the right, or the equation becomes invalid.
- Skipping the division step: if the equation starts as 2x² + 8x = 12, you must divide by 2 before computing half of b.
- Forgetting the ± sign: square-rooting both sides always gives two branches — dropping one loses half the solution.
- Adding instead of completing: make sure the value you add is exactly (b/2)²; guessing a round number does not create a perfect square.
Verify your answer by plugging each x-value back into the original equation. It takes ten seconds and catches nearly every arithmetic slip.