#### Answer

$\frac{-v_0\pm\sqrt{v_0^{2}+2gh}}{g}$

#### Work Step by Step

$h = \displaystyle \frac{1}{2}gt^{2}+v_0t$
$\frac{1}{2}gt^{2}+v_0t-h = 0$
We use the quadratic formula:
$t=\displaystyle \frac{-(v_0)\pm\sqrt{(v_0)^{2}-4(\frac{1}{2}g)(-h)}}{2(\frac{1}{2}g)}=\frac{-v_0\pm\sqrt{v_0^{2}+2gh}}{g}$