Answer
{$\frac{-1\pm2\sqrt 5}{4}$}
Work Step by Step
$(4x+1)^{2}=20$
Using generalized square root property, $4x+1=\pm\sqrt (20)$
$4x+1=\pm\sqrt (4\times5)$
$4x+1=\pm\sqrt 4\sqrt 5$
$4x+1=\pm2\sqrt 5$
Subtracting 1 from both sides, $4x+1-1=-1\pm2\sqrt 5$
$4x=-1\pm2\sqrt 5$
Dividing both sides by 4, $\frac{4x}{4}=\frac{-1\pm2\sqrt 5}{4}$
$x=\frac{-1\pm2\sqrt 5}{4}$
Check:
$(4(\frac{-1+2\sqrt 5}{4})+1)^{2}=20$
$(-1+2\sqrt 5+1)^{2}=20$
$(2\sqrt 5)^{2}=20$
$4(\sqrt 5)^{2}=20$
$4(5)=20$
$20=20$
$(4(\frac{-1-2\sqrt 5}{4})+1)^{2}=20$
$(-1-2\sqrt 5+1)^{2}=20$
$(-2\sqrt 5)^{2}=20$
$4(\sqrt 5)^{2}=20$
$4(5)=20$
$20=20$
Therefore, the solution set is {$\frac{-1\pm2\sqrt 5}{4}$}
