Answer
The solution of the equation $s=-16{{t}^{2}}+{{v}_{0}}t$ for $t$ is $t=\frac{{{v}_{0}}\pm \sqrt{v_{0}^{2}-64s}}{32}$.
Work Step by Step
Consider the equation $s=-16{{t}^{2}}+{{v}_{0}}t$.
Subtract $s$ from both sides.
$\begin{align}
& s-s=-16{{t}^{2}}+{{v}_{0}}t-s \\
& 0=-16{{t}^{2}}+{{v}_{0}}t-s \\
& -16{{t}^{2}}+{{v}_{0}}t-s=0
\end{align}$
Use the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve the equation $-16{{t}^{2}}+{{v}_{0}}t-s=0$ where $x=t$, $a=-16$, $b={{v}_{0}}$ and $c=-s$.
$\begin{align}
& t=\frac{-{{v}_{0}}\pm \sqrt{{{\left( {{v}_{0}} \right)}^{2}}-4\left( -16 \right)\left( -s \right)}}{2\left( -16 \right)} \\
& =\frac{-{{v}_{0}}\pm \sqrt{{{\left( {{v}_{0}} \right)}^{2}}-64s}}{-32} \\
& =\frac{{{v}_{0}}\pm \sqrt{v_{0}^{2}-64s}}{32}
\end{align}$
The solution of the equation is $t=\frac{{{v}_{0}}\pm \sqrt{v_{0}^{2}-64s}}{32}$.
