A quadratic like x^2 + bx + c rarely factorises into a neat perfect square on its own. Completing the square rewrites it so that it does contain a perfect square — a single squared bracket — plus a leftover number:

The squared bracket uses \tfrac{b}{2} — exactly half of the x-coefficient. The leftover number repairs the difference, because expanding \left(x + \tfrac{b}{2}\right)^2 gives an extra \left(\tfrac{b}{2}\right)^2 that we must subtract back off.

A worked example

Take x^2 + 6x + 5. Half of 6 is 3, so the square is (x + 3)^2. But (x + 3)^2 = x^2 + 6x + 9 — that is 4 too big compared with the +5 we started with, so we subtract 4:

Check it: (x + 3)^2 - 4 = x^2 + 6x + 9 - 4 = x^2 + 6x + 5. Both forms describe the same expression — one just wears its square on the outside.

Why "completing the square"?

The name is literal. Picture x^2 as a square of side x, and bx as a strip beside it. Cut the strip in half, lay the two \tfrac{b}{2}-wide halves along two sides of the square, and you almost have a bigger square of side x + \tfrac{b}{2} — except for a missing little corner of area \left(\tfrac{b}{2}\right)^2. Fill that corner in and the square is complete; since we added \left(\tfrac{b}{2}\right)^2, we must subtract it again to keep the value unchanged. Step through it.

The move builds straight on factorising harder quadratics: when a quadratic refuses to factorise into whole-number brackets, completing the square always works — it never needs the constant to cooperate. It is the engine behind the quadratic formula and the key to finding the turning point of a parabola.

See it explained

Khan Academy walks through completing the square step by step: