## College Algebra (11th Edition)

$x=\left\{ \dfrac{-1-2\sqrt{5}}{4},\dfrac{-1+2\sqrt{5}}{4} \right\}$
Taking the square root of both sides, the solutions to the given equation, $(4x+1)^2=20 ,$ are \begin{array}{l}\require{cancel} 4x+1=\pm\sqrt{20} \\\\ 4x+1=\pm\sqrt{4\cdot5} \\\\ 4x+1=\pm\sqrt{(2)^2\cdot5} \\\\ 4x+1=\pm2\sqrt{5} \\\\ 4x=-1\pm2\sqrt{5} \\\\ x=\dfrac{-1\pm2\sqrt{5}}{4} .\end{array} Hence, the solutions are $x=\left\{ \dfrac{-1-2\sqrt{5}}{4},\dfrac{-1+2\sqrt{5}}{4} \right\} .$