## Elementary Algebra

$x=\{ -2,1 \}$
Squaring both sides of the equation and using the properties of radicals, the solution to the given equation is \begin{array}{l}\require{cancel}\left( \sqrt{3x+6} \right)^2=\left( x+2 \right)^2 \\\\ 3x+6=(x)^2+2(x)(2)+(2)^2 \\\\ 3x+6=x^2+4x+4 \\\\ 0=x^2+(4x-3x)+(4-6) \\\\ x^2+x-2=0 \\\\ (x+2)(x-1)=0 \\\\ x=\{ -2,1 \} .\end{array} Upon checking, both solutions, $x=\{ -2,1 \} ,$ satisfy the original equation.