Answer
$r=\dfrac{5}{6}$
Work Step by Step
Squaring both sides of the given equation, $
r=\sqrt{\dfrac{20-19r}{6}}
,$ results to
\begin{align*}\require{cancel}
(r)^2&=\left(\sqrt{\dfrac{20-19r}{6}}\right)^2
\\\\
r^2&=\dfrac{20-19r}{6}
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}
6\cdot r^2&=\dfrac{20-19r}{\cancel6}\cdot\cancel6
\\\\
6r^2&=20-19r
\\
6r^2+19r-20&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(6r-5)(r+4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
6r-5=0 & r+4=0
\\
6r=5 & r=-4
\\\\
r=\dfrac{5}{6}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }r=\dfrac{5}{6}: & \text{If }r=-4
\\\\
\dfrac{5}{6}\overset{?}=\sqrt{\dfrac{20-19\left(\frac{5}{6}\right)}{6}} &
-4\overset{?}=\sqrt{\dfrac{20-19(-4)}{6}}
\\\\
\dfrac{5}{6}\overset{?}=\sqrt{\dfrac{20-\frac{95}{6}}{6}} &
-4\ne\text{ some nonnegative number}
\\\\
\dfrac{5}{6}\overset{?}=\sqrt{\dfrac{\frac{120}{6}-\frac{95}{6}}{6}}
\\\\
\dfrac{5}{6}\overset{?}=\sqrt{\dfrac{\frac{25}{6}}{6}}
\\\\
\dfrac{5}{6}\overset{?}=\sqrt{\dfrac{25}{36}}
\\\\
\dfrac{5}{6}\overset{\checkmark}=\dfrac{5}{6}
.\end{array}
Since $r=-4$ does not satisfy the original equation, then the only solution of the equation $
r=\sqrt{\dfrac{20-19r}{6}}
$ is $r=\dfrac{5}{6}$.
