Answer
$\left\{-4\right\}$
Work Step by Step
Squaring both sides, the given equation, $
-2r=\sqrt{\dfrac{48-20r}{2}}
$, is equivalent to
\begin{align*}
(-2r)^2&=\left(\sqrt{\dfrac{48-20r}{2}}\right)^2
\\\\
4r^2&=\dfrac{48-20r}{2}
.\end{align*}
Using properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
2(4r^2)&=\left(\dfrac{48-20r}{\cancel2}\right)\cancel2
\\\\
8r^2&=48-20r
\\
8r^2+20r-48&=0
\\\\
\dfrac{8r^2+20r-48}{4}&=\dfrac{0}{4}
\\\\
2r^2+5r-12&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(r+4)(2r-3)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
r+4=0 & 2r-3=0
\\
r=-4 & 2r=3
\\
& r=\dfrac{3}{2}
.\end{array}
Since both sides of the given radical equation were squared, then checking of solutions is a must for possible existence of extraneous solutions. Checking the solutions by substituting in the original equation results to
\begin{array}{l|r}
\text{If }r=-4 & \text{If }r=\dfrac{3}{2}:
\\\\
-2(-4)\overset{?}=\sqrt{\dfrac{48-20(-4)}{2}} & -2\left(\dfrac{3}{2}\right)\overset{?}=\sqrt{\dfrac{48-20\left(\dfrac{3}{2}\right)}{2}}
\\\\
8\overset{?}=\sqrt{\dfrac{48+80}{2}} & -3\overset{?}=\sqrt{\dfrac{48-30}{2}}
\\\\
8\overset{?}=\sqrt{\dfrac{128}{2}} & -3\ne\text{some nonnegative number.}
\\\\
8\overset{?}=\sqrt{64} &
\\
8\overset{\checkmark}=8 &
\end{array}
Since $r=\dfrac{3}{2}$ does not satisfy the original equation, then the only solution is $r=-4$. Hence, the solution set of the equation $
-2r=\sqrt{\dfrac{48-20r}{2}}
$ is $
\left\{-4\right\}
$.
