## College Algebra (10th Edition)

$\color{blue}{x=\left\{-2-2\sqrt2, -2+2\sqrt2\right\}}$
Square both sides: $(\sqrt{x^2+4x})^2=2^2 \\x^2+4x=4$ Complete the square by adding $(\frac{4}{2})^2=2^2=4$ 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 solutions are: $\color{blue}{x=\left\{-2-2\sqrt2, -2+2\sqrt2\right\}}$