How to Complete the Square: Step-by-Step Method with Examples

To complete the square on an expression like x² + bx, take half of the coefficient b, square it, and add that number to create a perfect-square trinomial. That trinomial factors as (x + b/2)². This trick rewrites quadratics into vertex form, solves equations, and converts general conic equations into standard form.

Below is the method demonstrated on two full problems, plus the mistakes that trip most students up.

A first worked example: solve x² + 6x − 7 = 0

We'll solve x² + 6x − 7 = 0 by completing the square.

1. Move the constant. Add 7 to both sides: x² + 6x = 7.
2. Take half of the middle coefficient. Half of 6 is 3.
3. Square it. 3² = 9.
4. Add 9 to both sides. x² + 6x + 9 = 7 + 9, which gives x² + 6x + 9 = 16.
5. Factor the perfect square. The left side is (x + 3)², so (x + 3)² = 16.
6. Take the square root of both sides. x + 3 = ±4.
7. Solve. x = −3 + 4 = 1 or x = −3 − 4 = −7.

The positive solution is x = 1. (The other root is x = −7, which you can verify directly: (−7)² + 6(−7) − 7 = 49 − 42 − 7 = 0.)

The completing-the-square formula and why it works

The identity behind the method is:

x² + bx + (b/2)² = (x + b/2)²

Any quadratic x² + bx is one term away from being a perfect square. That missing term is (b/2)². Adding it (and adjusting the other side of the equation to keep things balanced) is what "completing the square" means.

The step-by-step procedure

1. Make the leading coefficient 1. If your quadratic is 2x² − 12x + 16, factor 2 out of the x-terms first: 2(x² − 6x) + 16.
2. Isolate the x-terms. Move the constant to the other side of the equation, or set it aside inside the parentheses.
3. Compute (b/2)². For x² − 6x, half of −6 is −3, and (−3)² = 9.
4. Add and subtract (or add to both sides). Keep the equation balanced. Inside a factored expression like 2(x² − 6x + 9 − 9), you add and subtract the same number.
5. Factor the perfect-square trinomial. x² − 6x + 9 = (x − 3)². Now the expression is in vertex or standard conic form.

A second worked example: rewrite y = x² − 8x + 11 in vertex form

Vertex form is y = (x − h)² + k. Let's convert y = x² − 8x + 11.

1. Group the x-terms. y = (x² − 8x) + 11.
2. Half of −8 is −4; squared is 16. That's the number we need.
3. Add and subtract 16 inside. y = (x² − 8x + 16) − 16 + 11.
4. Factor the perfect square. y = (x − 4)² − 5.

The vertex is (4, −5). To double-check, if we solve x² − 8x + 11 = −5, we get (x − 4)² = 0, so x = 4 — exactly the vertex's x-coordinate.

Common mistakes when completing the square

• Forgetting to divide b by 2 before squaring. Students often square the whole coefficient. For x² + 6x, the number you add is 9 (from (6/2)²), not 36.
• Not balancing the equation. If you add 9 to the left side of an equation, you must add 9 to the right side too — or subtract 9 on the same side.
• Skipping the leading-coefficient step. If the x² term has a coefficient other than 1, you must factor it out of the x-terms first. Otherwise the (b/2)² formula gives the wrong number.
• Sign errors on the vertex. Vertex form (x − h)² means h has the opposite sign of what's inside. (x + 3)² has vertex x-coordinate −3, not +3.
• Losing the ± when taking the square root. A squared expression equal to a positive number always yields two solutions.

Ready to practice this yourself?

Completing the square looks mechanical once you've done a few, but the sign traps and the leading-coefficient step catch almost everyone the first time through. LernOS's guided tutor walks you through problems one prompt at a time — it asks what step comes next instead of just showing the answer, so the method actually sticks. Sign up below to try it on your own quadratic.