#### Answer

$s=9+4\sqrt{5}$

#### Work Step by Step

$ s-4\sqrt{s}-1=0\qquad$...substitute $\sqrt{s}$ for $u$ so that $u^{2}=s$
Note that $s$ and $\sqrt{s}$ are positive.
.
$ u^{2}-4u-1=0\qquad$... solve with the Quadractic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{4\pm\sqrt{16+4}}{2}$
$u=\displaystyle \frac{4\pm\sqrt{20}}{2}$
$u=\displaystyle \frac{4\pm 2\sqrt{5}}{2}$
$u=\displaystyle \frac{4+2\sqrt{5}}{2}=2+\sqrt{5}$ or $u=\displaystyle \frac{4-2\sqrt{5}}{2}=2-2\sqrt{5}$
Bring back $\sqrt{s}=u$.
$\sqrt{s}=2+\sqrt{5}$ or $\sqrt{s}=2-2\sqrt{5}$
$\sqrt{s}$ is a positive number which is why we discard $2-2\sqrt{5}$.
$s=(2+\sqrt{5})^{2}=4+4\sqrt{5}+5=9+4\sqrt{5}$