Answer
$x=-1$
Work Step by Step
Squaring both sides of the given equation, $
-x=\sqrt{\dfrac{3x+7}{4}}
,$ results to
\begin{align*}\require{cancel}
(-x)^2&=\left(\sqrt{\dfrac{3x+7}{4}}\right)^2
\\\\
x^2&=\dfrac{3x+7}{4}
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}
4\cdot x^2&=\dfrac{3x+7}{\cancel4}\cdot\cancel4
\\\\
4x^2&=3x+7
\\
4x^2-3x-7&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(x+1)(4x-7)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x+1=0 & 4x-7=0
\\
x=-1 & 4x=7
\\\\
& x=\dfrac{7}{4}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=-1: & \text{If }x=\dfrac{7}{4}
\\\\
-(-1)\overset{?}=\sqrt{\dfrac{3(-1)+7}{4}} &
-\dfrac{7}{4}\overset{?}=\sqrt{\dfrac{3\left(\frac{7}{4}\right)+7}{4}}
\\\\
1\overset{?}=\sqrt{\dfrac{-3+7}{4}} &
-\dfrac{7}{4}\overset{?}=\sqrt{\dfrac{\frac{21}{4}+7}{4}}
\\\\
1\overset{?}=\sqrt{\dfrac{4}{4}} &
-\dfrac{7}{4}\overset{?}=\sqrt{\dfrac{\frac{21}{4}+\frac{28}{4}}{4}}
\\\\
1\overset{\checkmark}=1 &
-\dfrac{7}{4}\overset{?}=\sqrt{\dfrac{\frac{49}{4}}{4}}
\\\\
&
-\dfrac{7}{4}\overset{?}=\sqrt{\dfrac{49}{16}}
\\\\
&
-\dfrac{7}{4}\ne\dfrac{7}{4}
.\end{array}
Since $x=\dfrac{7}{4}$ does not satisfy the original equation, then the only solution of the equation $
-x=\sqrt{\dfrac{3x+7}{4}}
$ is $x=-1$.
