Answer
$r=-\dfrac{2pc}{3}\text{ or }r=\dfrac{5pc}{4}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
10p^2c^2+7pcr=12r^2
,$ is equivalent to
\begin{align*}\require{cancel}
0&=12r^2-7pcr-10p^2c^2
\\
12r^2-7pcr-10p^2c^2&=0
\\
(12)r^2-(7pc)r-10p^2c^2&=0
.\end{align*}
The equation above has
\begin{align*}
a=
12
,\text{ }b=
-7pc
,\text{ and }c=
-10p^2c^2
.\end{align*}
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
r&=
\dfrac{-(-7pc)\pm\sqrt{(-7pc)^2-4(12)(-10p^2c^2)}}{2(12)}
\\\\&=
\dfrac{7pc\pm\sqrt{49p^2c^2+480p^2c^2}}{24}
\\\\&=
\dfrac{7pc\pm\sqrt{529p^2c^2}}{24}
\\\\&=
\dfrac{7pc\pm\sqrt{529}\cdot\sqrt{(pc)^2}}{24}
\\\\&=
\dfrac{7pc\pm23|pc|}{24}
.\end{align*}
Assuming that $p$ and $c$ are both positive or that $p$ and $c$ are both negative, then
\begin{align*}
r&=\dfrac{7pc\pm23pc}{24}
\end{align*}\begin{array}{l|r}
r=\dfrac{7pc-23pc}{24} & r=\dfrac{7pc+23pc}{24}
\\\\
r=\dfrac{-16pc}{24} & r=\dfrac{30pc}{24}
\\\\
r=-\dfrac{2pc}{3} & r=\dfrac{5pc}{4}
\end{array}
Hence, in terms of $r,$ the given equation is equivalent to $r=-\dfrac{2pc}{3}\text{ or }r=\dfrac{5pc}{4}$.
