Question 264754
{{{(3x-3)^2 = 81)}}}
You need to take the square root of both sides,
which you tried to do, but went astray.
If something is squared, then taking the
square root exactly cancels out the squaring.
In general:
{{{sqrt(n^2) = n}}}, so
{{{sqrt((3x-3)^2) = sqrt(81)}}}
{{{3x - 3 = +9}}}
and
{{{3x - 3 = -9}}}
From the 1st, I get
{{{3x = 12}}}
{{{x = 4}}}
From the 2nd, I get
{{{3x = -6}}}
{{{x = -2}}}
You can check these answers by plugging the number into 
the original equation:
{{{(3x-3)^2 = 81)}}}
{{{(3*4 - 3)^2 = 81}}}
{{{9^2 = 81}}}
{{{81 = 81}}}
and
{{{(3x-3)^2 = 81)}}}
{{{(3*(-2) - 3)^2 = 81)}}}
{{{(-9)^2 = 81}}}
{{{81 = 81}}}
OK