Answer
$r=25$
Work Step by Step
Let $z=
1+\sqrt{r}
$. Then the given equation, $
2\left(1+\sqrt{r}\right)^2=13\left(1+\sqrt{r}\right)-6
,$ is equivalent to
\begin{align*}
2z^2&=13z-6
\\
2z^2-13z+6=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-6)(2z-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-6=0 & 2z-1=0
\\
z=6 & 2z=1
\\\\
& z=\dfrac{1}{2}
.\end{array}
Since $z=1+\sqrt{r}$, by back-substitution,
\begin{array}{l|r}
1+\sqrt{r}=6 & 1+\sqrt{r}=\dfrac{1}{2}
\\\\
\sqrt{r}=6-1 & \sqrt{r}=\dfrac{1}{2}-1
\\\\
\sqrt{r}=5 & \sqrt{r}=-\dfrac{1}{2}
\\\\
\left(\sqrt{r}\right)^2=(5)^2 & \left(\sqrt{r}\right)^2=\left(-\dfrac{1}{2}\right)^2
\\\\
r=25 & r=\dfrac{1}{4}
.\end{array}
Checking by substituting the solutions in the given equation results to
\begin{array}{l|r}
\text{If }r=25: & \text{If }r=\dfrac{1}{4}:
\\\\
2\left(1+\sqrt{25}\right)^2\overset{?}=13\left(1+\sqrt{25}\right)-6 &
2\left(1+\sqrt{\dfrac{1}{4}}\right)^2\overset{?}=13\left(1+\sqrt{\dfrac{1}{4}}\right)-6
\\\\
2\left(1+5\right)^2\overset{?}=13\left(1+5\right)-6 &
2\left(1+\dfrac{1}{2}\right)^2\overset{?}=13\left(1+\dfrac{1}{2}\right)-6
\\\\
2\left(6\right)^2\overset{?}=13\left(6\right)-6 &
2\left(\dfrac{3}{2}\right)^2\overset{?}=13\left(\dfrac{3}{2}\right)-6
\\\\
2\left(36\right)\overset{?}=78-6 &
2\left(\dfrac{9}{4}\right)\overset{?}=\dfrac{39}{2}-6
\\\\
72\overset{\checkmark}=72 &
\dfrac{9}{2}\overset{?}=\dfrac{39}{2}-\dfrac{12}{2}
\\\\
&
\dfrac{9}{2}\ne\dfrac{27}{2}
.\end{array}
Since $r=\dfrac{1}{4}$ does not satisfy the original equation then the only solution of the equation $
2\left(1+\sqrt{r}\right)^2=13\left(1+\sqrt{r}\right)-6
$ is $r=25$.
