Question 104542
{{{x^2-16x+63=0}}}
To complete the square, you want to find the form
{{{(x+a)^2=x^2+2ax+a^2}}}
By comparing the expanded perfect square equation to your equation, you see that,
{{{2ax=-16x}}}
{{{a=-8}}}
{{{a^2=64}}}
I have 63, so I just need to add 1, to have a perfect square form on the left hand side. 
{{{x^2-16x+63+1=1}}}
{{{x^2-16x+64=1}}}
{{{(x-8)^2=1}}}
{{{sqrt((x-8)^2)=sqrt(1)}}}
Since {{{sqrt(1)}}}=1 and {{{sqrt(1)}}}=-1, we will solve for two solutions. 
First,
{{{x-8=1}}}
{{{x=9}}}
Secnd
{{{x-8=-1}}}
{{{x=7}}}
Let's verify the solutions. 
{{{x^2-16x+63=0}}}
{{{9^2-16(9)+63=0}}}
{{{81-144+63=0}}}
{{{0=0}}}
Good answer.
{{{x^2-16x+63=0}}}
{{{7^2-16(7)+63=0}}}
{{{49-112+63=0}}}
{{{0=0}}}
Good answer.