Answer
The solution set is $\color{blue}{\left\{-2-2\sqrt2, -2+2\sqrt2\right\}}$.
Work Step by Step
Square both sides:
$(\sqrt{x^2+4x})^2=2^2
\\x^2+4x=4$
Complete the square by adding $\left(\frac{4}{2}\right)^2=2^2=4$ to both sides of the equation:
$x^2+4x+4=4+4
\\x^2+4x+4=8$
Factor the trinomial to obtain:
$(x+2)^2=8$
Take the square root of both sides:
$\sqrt{(x+2)^2} = \pm \sqrt{8}
\\x+2=\pm \sqrt{4\cdot2}
\\x+2 = \pm \sqrt{2^2\cdot 2}
\\x+2=\pm 2\sqrt{2}$
Subtract $2$ to both sides of the equation:
$x+2-2=-2 \pm 2\sqrt{2}
\\x=-2\pm2\sqrt{2}$
Thus, the solution set is $\color{blue}{\left\{-2-2\sqrt2, -2+2\sqrt2\right\}}$.
