In elementary algebra, completing the square is a technique for converting a quadratic polynomial of the form ⁠${\displaystyle \textstyle ax^{2}+bx+c}$⁠ to the form ⁠${\displaystyle \textstyle a(x-h)^{2}+k}$⁠ for some values of ⁠${\displaystyle h}$⁠ and ⁠${\displaystyle k}$⁠. In terms of a new quantity ⁠${\displaystyle x-h}$⁠, this expression is a quadratic polynomial with no linear term. By subsequently isolating ⁠${\displaystyle \textstyle (x-h)^{2}}$⁠ and taking the square root, a quadratic problem can be reduced to a linear problem.

The name completing the square comes from a geometrical picture in which ⁠${\displaystyle x}$⁠ represents an unknown length. Then the quantity ⁠${\displaystyle \textstyle x^{2}}$⁠ represents the area of a square of side ⁠${\displaystyle x}$⁠ and the quantity ⁠${\displaystyle {\tfrac {b}{a}}x}$⁠ represents the area of a pair of congruent rectangles with sides ⁠${\displaystyle x}$⁠ and ⁠${\displaystyle {\tfrac {b}{2a}}}$⁠. To this square and pair of rectangles, one more square is added, of side length ⁠${\displaystyle {\tfrac {b}{2a}}}$⁠. This crucial step completes a larger square of side length ⁠${\displaystyle x+{\tfrac {b}{2a}}}$⁠.

Completing the square is the oldest method of solving general quadratic equations, used in Old Babylonian clay tablets dating from 1800–1600 BCE, and is still taught in elementary algebra courses today. It is also used for graphing quadratic functions, deriving the quadratic formula, and more generally in computations involving quadratic polynomials, for example in calculus evaluating Gaussian integrals with a linear term in the exponent, and finding Laplace transforms.

The technique of completing the square was known in the Old Babylonian Empire.

Muhammad ibn Musa Al-Khwarizmi, a famous polymath who wrote the early algebraic treatise Al-Jabr, used the technique of completing the square to solve quadratic equations.

The formula in elementary algebra for computing the square of a binomial is: $${\displaystyle (x+p)^{2}\,=\,x^{2}+2px+p^{2}.}$$

For example: $${\displaystyle {\begin{alignedat}{2}(x+3)^{2}\,&=\,x^{2}+6x+9&&(p=3)\\[3pt](x-5)^{2}\,&=\,x^{2}-10x+25\qquad &&(p=-5).\end{alignedat}}}$$

In any perfect square, the coefficient of x is twice the number p, and the constant term is equal to p².