Question 1182190
The form and style of the question look very familiar.

If you wish to solve a quadratic equation using Completing the Square then for  {{{x^2+bx+c=0}}}, the term which does this is  {{{(b/2)^2=b^2/4}}}.


{{{x^2+bx=-c}}}

{{{x^2+bx+(b/2)^2=(b/2)^2-c}}}

{{{(x+b/2)^2=b^2/4-c}}}

etcetera, etcetera until you finish solve for x, and simplify.
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Your example given is {{{x^2-2x-12=0}}}, so

{{{x^2-2x=12}}}

term you need is 1.

{{{x^2-2x+1=12+1}}}

{{{x^2-2x+1=13}}}

{{{(x-1)^2=13}}}

{{{x-1=0+- sqrt(13)}}}

{{{x=1+- sqrt(13)}}}