Answer
$t=\dfrac{-v_{0}\pm\sqrt{v^{2}_{0}+2gh}}{g}$
Work Step by Step
$h=\dfrac{1}{2}gt^{2}+v_{0}t$; for $t$
Take $h$ to the right side of the equation to set it equal to $0$:
$\dfrac{1}{2}gt^{2}+v_{0}t-h=0$
Multiply the whole equation by $2$, to avoid working with fractions:
$gt^{2}+2v_{0}t-2h=0$
This is a quadratic equation. Let's solve it using the quadratic formula, which is:
$t=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
For this particular equation, $a=g$, $b=2v_{0}$ and $c=-2h$
Substitute:
$t=\dfrac{-2v_{0}\pm\sqrt{(2v_{0})^{2}-4g(-2h)}}{2g}$
Simplify:
$t=\dfrac{-2v_{0}\pm\sqrt{4v^{2}_{0}+8gh}}{2g}$
$t=\dfrac{-2v_{0}\pm\sqrt{4(v^{2}_{0}+2gh)}}{2g}$
$t=\dfrac{-2v_{0}\pm2\sqrt{v^{2}_{0}+2gh}}{2g}$
$t=\dfrac{-v_{0}\pm\sqrt{v^{2}_{0}+2gh}}{g}$
