#### Answer

$t=\pm\dfrac{\sqrt{dk }}{k}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
d=kt^2
,$ in terms of $
t
,$ use the properties of equality and the Square Root Principle to isolate the variable.
$\bf{\text{Solution Details:}}$
Using the properties of equality, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{d}{k}=t^2
\\\\
t^2=\dfrac{d}{k}
.\end{array}
Taking the square root of both sides (Square Root Principle), the equation above is equivalent to
\begin{array}{l}\require{cancel}
t=\pm\sqrt{\dfrac{d}{k}}
.\end{array}
Rationalizing the denominator by 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{d}{k}\cdot\dfrac{k}{k}}
\\\\
t=\pm\sqrt{\dfrac{dk}{k^2}}
\\\\
t=\pm\sqrt{\dfrac{1}{k^2}\cdot dk }
\\\\
t=\pm\sqrt{\left(\dfrac{1}{k} \right)^2\cdot dk }
\\\\
t=\pm\dfrac{1}{k}\sqrt{dk }
\\\\
t=\pm\dfrac{\sqrt{dk }}{k}
.\end{array}