Question 28505
{{{(2x+2)^2 = 64 }}}

Not quite...

Remember, squaring a binomial isn't the same as squaring each term.

{{{(2x+2)^2<>4x^2+4}}} it's actually the same as multiplying the two terms together:

{{{(2x+2)^2=(2x+2)(2x+2)=4x^2+8x+4}}}

Anyway, that's not getting you any closer to the answer, the correct way to approach this problem is to take the square root of both sides.

{{{(2x+2)^2=64}}} 

When you take the square root, you have to remember that you'll actually have two equations, because 2x+2 could be -8 or 8 and either way when you square it you'll get 64, so we write:
{{{2x+2=0+- 8}}} (ignore the 0, it's just to get the +- symbol to work in the equation editor)
Or you could write it as two equations {{{2x+2=8}}} and {{{2x+2=-8}}}

When you subtract 2 from both sides you get:
{{{2x=-2+- 8}}} Divide by 2:
{{{x=-1+- 4}}}
So your answers are {{{x=-1+4=3}}} and {{{x=-1-4=5}}}
x=3, -5

Graphically (green = (2x+2)^2 & red = 64) crosses at x=3 and x=-5:
{{{graph(300, 300, -10, 10, -10, 80, 64, (2x+2)^2)}}}