#### Answer

$v=\pm\dfrac{\sqrt{kAF}}{F}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
F=\dfrac{kA}{v^2}
,$ in terms of $
v
,$ use the properties of equality and the Square Root Principle to isolate the variable.
$\bf{\text{Solution Details:}}$
Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to
\begin{array}{l}\require{cancel}
F(v^2)=1(kA)
\\\\
Fv^2=kA
\\\\
v^2=\dfrac{kA}{F}
.\end{array}
Taking the square root of both sides (Square Root Principle), the equation above is equivalent to
\begin{array}{l}\require{cancel}
v=\pm\sqrt{\dfrac{kA}{F}}
.\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}
v=\pm\sqrt{\dfrac{kA}{F}\cdot\dfrac{F}{F}}
\\\\
v=\pm\sqrt{\dfrac{kAF}{F^2}}
\\\\
v=\pm\sqrt{\dfrac{1}{F^2}\cdot kAF}
\\\\
v=\pm\sqrt{\left( \dfrac{1}{F} \right)^2\cdot kAF}
\\\\
v=\pm\dfrac{1}{F}\sqrt{kAF}
\\\\
v=\pm\dfrac{\sqrt{kAF}}{F}
.\end{array}