#### Answer

$t=\pm\dfrac{\sqrt{g(s-s_0-k)}}{g}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the properties of equality to solve the given equation, $
s=s_0+gt^2+k
,$ for $
t
.$
$\bf{\text{Solution Details:}}$
Using the properties of equality, the equation above is equivalent to
\begin{array}{l}\require{cancel}
s-s_0-k=gt^2
\\\\
\dfrac{s-s_0-k}{g}=t^2
\\\\
t^2=\dfrac{s-s_0-k}{g}
.\end{array}
Taking the square root of both sides (Square Root Principle) results to
\begin{array}{l}\require{cancel}
t=\pm\sqrt{\dfrac{s-s_0-k}{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{s-s_0-k}{g}\cdot\dfrac{g}{g}}
\\\\
t=\pm\sqrt{\dfrac{g(s-s_0-k)}{g^2}}
\\\\
t=\pm\sqrt{\dfrac{1}{g^2}\cdot g(s-s_0-k)}
\\\\
t=\pm\sqrt{\left( \dfrac{1}{g} \right)^2 \cdot g(s-s_0-k)}
\\\\
t=\pm\left| \dfrac{1}{g} \right|\sqrt{g(s-s_0-k)}
.\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{g(s-s_0-k)}
\\\\
t=\pm\dfrac{\sqrt{g(s-s_0-k)}}{g}
.\end{array}