## Precalculus (6th Edition)

$\color{blue}{\left\{\dfrac{-1-2\sqrt3}{4}, \dfrac{-1+2\sqrt3}{4}\right\}}$
RECALL: If $x^2=a$, then taking the square root of both sides gives $x = \pm \sqrt{a}$. Take the square root of both sides of the given equation to obtain: $\sqrt{(4x+1)^2}=\pm \sqrt{20} \\4x+1 =\pm \sqrt{4(5)} \\4x+1=\pm\sqrt{2^2(5)} \\4x+1 =\pm 2\sqrt{5}$ Subtract $1$ to both sides: $4x =-1 \pm 2\sqrt{5}$ Divide $4$ to both sides: $x=\dfrac{-1\pm2\sqrt3}{4}$ Thus, the solution set is $\color{blue}{\left\{\dfrac{-1-2\sqrt3}{4}, \dfrac{-1+2\sqrt3}{4}\right\}}$.