#### Answer

$\color{blue}{\left\{\dfrac{-1-2\sqrt3}{4}, \dfrac{-1+2\sqrt3}{4}\right\}}$

#### Work Step by Step

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 $\color{blue}{\left\{\dfrac{-1-2\sqrt3}{4}, \dfrac{-1+2\sqrt3}{4}\right\}}$.