Answer
$t=\pm\dfrac{\sqrt{2sg}}{g}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the properties of equality to solve the given equation, $
s=\dfrac{1}{2}gt^2
,$ for $
t
.$
$\bf{\text{Solution Details:}}$
Multiplying both sides by $
2
,$ and then dividing by $
g
$ result to
\begin{array}{l}\require{cancel}
2s=gt^2
\\\\
\dfrac{2s}{g}=t^2
\\\\
t^2=\dfrac{2s}{g}
.\end{array}
Taking the square root of both sides (Square Root Principle) results to
\begin{array}{l}\require{cancel}
t=\pm\sqrt{\dfrac{2s}{g}}
.\end{array}
Multiplying the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index results to
\begin{array}{l}\require{cancel}
t=\pm\sqrt{\dfrac{2s}{g}\cdot\dfrac{g}{g}}
\\\\
t=\pm\sqrt{\dfrac{2sg}{g^2}}
\\\\
t=\pm\sqrt{\dfrac{1}{g^2}\cdot2sg}
\\\\
t=\pm\sqrt{\left( \dfrac{1}{g} \right)^2\cdot2sg}
\\\\
t=\pm\left| \dfrac{1}{g} \right|\sqrt{2sg}
.\end{array}
Assuming that all variables are positive, the equation above is equivalent to
\begin{array}{l}\require{cancel}
t=\pm\dfrac{1}{g}\sqrt{2sg}
\\\\
t=\pm\dfrac{\sqrt{2sg}}{g}
.\end{array}