Question 77475
*[invoke completing_the_square 1, -6, 8]

So the quadratic {{{y=x^2 - 6x + 8}}} can be converted to the standard form {{{y=a (x - h)^2 + k}}} which is

{{{y=(x-3)^2-1}}} where a=1, h=3, and k=-1

Now we let y=0

{{{0=(x-3)^2-1}}}

{{{1=(x-3)^2}}} Add 1 to both sides

{{{sqrt(1)=sqrt((x-3)^2)}}}

*[Tex \large x-3=\pm 1]

Which means 


{{{x-3=1}}} or {{{x-3=-1}}}

Now add 3 to both sides for each case to solve for x:

So our solutions are

{{{x=4}}} or {{{x=2}}}