## Elementary Algebra

$y=\{ -5,-4\}$
Squaring both sides of the inequality and using the properties of radicals, the solution to the given equation is \begin{array}{l}\require{cancel}\left( \sqrt{y+5} \right)^2=\left( y+5 \right)^2 \\\\ y+5=(y)^2+2(y)(5)+(5)^2 \\\\ y+5=y^2+10y+25 \\\\ 0=y^2+(10y-y)+(25-5) \\\\ y^2+9y+20=0 \\\\ (y+4)(y+5)=0 \\\\ y=\{ -5,-4\} .\end{array} Upon checking, both solutions, $y=\{ -5,-4\} ,$ satisfy the original equation.