Question 87586
{{{(2x+1)^2=36}}}


{{{2x+1=sqrt(36)}}} Take the square root of both sides


*[Tex \LARGE 2x+1=\pm 6] Take the square root of 36


So it breaks down to 


{{{2x+1=6}}} or {{{2x+1=-6}}}


{{{2x+1=6-1}}} or {{{2x+1=-6-1}}} Subtract 1 from both sides for each case


{{{2x=5}}} or {{{2x=-7}}}


{{{x=5/2}}} or {{{x=-7/2}}} Divide both sides by 2. So this is our answer