Answer
$x=-2$
Work Step by Step
Squaring both sides of the given equation, $
-x=\sqrt{\dfrac{8-2x}{3}}
,$ results to
\begin{align*}\require{cancel}
(-x)^2&=\left(\sqrt{\dfrac{8-2x}{3}}\right)^2
\\\\
x^2&=\dfrac{8-2x}{3}
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}
3\cdot x^2&=\dfrac{8-2x}{\cancel3}\cdot\cancel3
\\\\
3x^2&=8-2x
\\
3x^2+2x-8&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(x+2)(3x-4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x+2=0 & 3x-4=0
\\
x=-2 & 3x=4
\\\\
& x=\dfrac{4}{3}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=-2: & \text{If }x=\dfrac{4}{3}
\\\\
-(-2)\overset{?}=\sqrt{\dfrac{8-2(-2)}{3}} &
-\dfrac{4}{3}\overset{?}=\sqrt{\dfrac{8-2\left(\frac{4}{3}\right)}{3}}
\\\\
2\overset{?}=\sqrt{\dfrac{8+4}{3}} &
-\dfrac{4}{3}\ne\text{ some nonnegative number}
\\\\
2\overset{?}=\sqrt{\dfrac{12}{3}}
\\\\
2\overset{?}=\sqrt{4}
\\
2\overset{\checkmark}=2
.\end{array}
Since $x=\dfrac{4}{3}$ does not satisfy the original equation, then the only solution of the equation $
-x=\sqrt{\dfrac{8-2x}{3}}
$ is $x=-2$.
