#### Answer

$t=\pm\dfrac{\sqrt{2a(r-r_0)}}{a}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the properties of equality to solve the given equation, $
r=r_0+\dfrac{1}{2}at^2
,$ for $
t
.$
$\bf{\text{Solution Details:}}$
Using the properties of equality, the equation above is equivalent to
\begin{array}{l}\require{cancel}
r-r_0=\dfrac{1}{2}at^2
\\\\
2(r-r_0)=\left( \dfrac{1}{2}at^2 \right)2
\\\\
2(r-r_0)=at^2
\\\\
\dfrac{2(r-r_0)}{a}=t^2
\\\\
t^2=\dfrac{2(r-r_0)}{a}
.\end{array}
Taking the square root of both sides (Square Root Principle) results to
\begin{array}{l}\require{cancel}
t=\pm\sqrt{\dfrac{2(r-r_0)}{a}}
.\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{2(r-r_0)}{a}\cdot\dfrac{a}{a}}
\\\\
t=\pm\sqrt{\dfrac{2a(r-r_0)}{a^2}}
\\\\
t=\pm\sqrt{\dfrac{1}{a^2}\cdot2a(r-r_0)}
\\\\
t=\pm\sqrt{\left( \dfrac{1}{a} \right)^2\cdot2a(r-r_0)}
\\\\
t=\pm\left| \dfrac{1}{a} \right|\sqrt{2a(r-r_0)}
.\end{array}
Assuming that all variables are positive, the equation above is equivalent to
\begin{array}{l}\require{cancel}
t=\pm\dfrac{1}{a}\sqrt{2a(r-r_0)}
\\\\
t=\pm\dfrac{\sqrt{2a(r-r_0)}}{a}
.\end{array}