#### Answer

$h=\pm\dfrac{d^2\sqrt{kL}}{L}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
L=\dfrac{kd^4}{h^2}
,$ in terms of $
h
,$ 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}
L(h^2)=1(kd^4)
\\\\
Lh^2=kd^4
\\\\
h^2=\dfrac{kd^4}{L}
.\end{array}
Taking the square root of both sides (Square Root Principle), the equation above is equivalent to
\begin{array}{l}\require{cancel}
h=\pm\sqrt{\dfrac{kd^4}{L}}
.\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}
h=\pm\sqrt{\dfrac{kd^4}{L}\cdot\dfrac{L}{L}}
\\\\
h=\pm\sqrt{\dfrac{kd^4L}{L^2}}
\\\\
h=\pm\sqrt{\dfrac{d^4}{L^2}\cdot kL}
\\\\
h=\pm\sqrt{\left( \dfrac{d^2}{L} \right)^2\cdot kL}
\\\\
h=\pm\dfrac{d^2}{L}\sqrt{kL}
\\\\
h=\pm\dfrac{d^2\sqrt{kL}}{L}
.\end{array}