## Elementary Algebra

$x=\{-3,-2\}$
Squaring both sides of the equation and then using the properties of equality, we obtain: \begin{array}{l}\require{cancel}\left( \sqrt{x+3} \right)^2=\left( x+3 \right)^2 \\\\ x+3=(x)^2+2(x)(3)+(3)^2 \\\\ x+3=x^2+6x+9 \\\\ 0=x^2+(6x-x)+(9-3) \\\\ x^2+5x+6=0 \\\\ (x+3)(x+2)=0 \\\\ x=\{-3,-2\} .\end{array} Upon checking, both solutions satisfy the original equation. Hence, $x=\{-3,-2\} .$