Question 55250
Solve {{{(3x+3)^2=16}}} 
Use the square root property:
For any real number c, if {{{x^2=c}}}, then {{{x=sqrt(c)}}} or {{{x=-sqrt(c)}}}.
{{{(3x+3)^2=16}}}
{{{sqrt((3x+3)^2)=+-sqrt(16)}}}
{{{3x+3=4}}} or {{{3x+3=-4}}}
{{{3x+3-3=4-3}}} or {{{3x+3-3=-4-3}}}
{{{3x=1}}} or {{{3x=-7}}}
{{{3x/3=1/3}}} or {{{3x/3=-7/3}}}
{{{highlight(x=1/3)}}} or {{{highlight(x=-7/3)}}}
:
Check:
if x=1/3
{{{(3(1/3)+3)^2=16}}}
{{{(1+3)^2=16}}}
{{{4^2=16}}}
{{{16=16}}}  x=1/3 checks out.
if x=-7/3
{{{(3(-7/3)+3)^2=16}}}
{{{(-7+3)^2=16}}}
{{{(-4)^2=16}}}
{{{16=16}}}  x=-7/3 also check out, so we're right!!!
Happy Calculating!!!