Question 856704
You are correct. You would square both sides to eliminate the square root. Here is one way to do it.



{{{24 = sqrt(x+2)sqrt(3x-6)}}}



{{{24 = sqrt((x+2)(3x-6))}}}  Look at Rule 1 under the "distributing" section from <a href="http://www.mathwords.com/s/square_root_rules.htm">this page</a>



{{{24^2 = (sqrt((x+2)(3x-6)))^2}}} Square both sides



{{{576 = (x+2)(3x-6)}}}



{{{576 = x(3x-6)+2(3x-6)}}}



{{{576 = 3x^2-6x+6x-12}}}



{{{576 = 3x^2-12}}}



{{{3x^2-12 = 576}}}



{{{3x^2 = 576+12}}}



{{{3x^2 = 588}}}



{{{x^2 = 588/3}}}



{{{x^2 = 196}}}



{{{x = ""+-sqrt(196)}}}



{{{x = sqrt(196)}}} or {{{x = -sqrt(196)}}}



{{{x = 14}}} or {{{x = -14}}}



So the <i>possible</i> solutions are {{{x = 14}}} or {{{x = -14}}}



I'll leave the check to you to do. You need to plug each possible answer into the original equation and simplify. If you get a true statement, then the possible solution is a real solution. If not, then it's not a real solution.