To solve the equation #x^2+2x−63=0#, using the completing the square method, one has to complete the square using the variable #x#.

We know the identity #(x+a)^2=x^2+2ax+a^2#, hence we should halve the coefficient of #x# and square it and then add and subtract the 'square' in the trinomial. Here coefficient of #x# is 2 and hence squareof half of it is #1#. Hence

#x^2+2x−63=0# can be written as

#x^2+2x+1-1−63=0# or

#(x^2+2xxx xx1+1^2)-64# or

#(x+1)^2-8^2#

Now using the identity #a^2-b^2=(a+b)(a-b)#, this becomes

#((x+1)+8)xx((x+1)-8)# or

#(x+9)(x-7)#