Answer
$x=\{ -1,4 \}$
Work Step by Step
Let $z=
x^2-3x
.$ Then the given equation, $
7-(x^2-3x)=\sqrt{(x^2-3x)+5}
$, is equivalent to
\begin{array}{l}\require{cancel}
7-z=\sqrt{z+5}
.\end{array}
Squaring both sides of the equal sign results to
\begin{array}{l}\require{cancel}
(7)^2+2(7)(-z)+(-z)^2=z+5
\\\\
49-14z+z^2=z+5
\\\\
z^2+(-14z-z)+(49-5)=0
\\\\
z^2-15z+44=0
\\\\
(z-11)(z-4)=0
\\\\
z=\left\{ 4,11 \right\}
.\end{array}
If $z=4$, then,
\begin{array}{l}\require{cancel}
x^2-3x=4
\\\\
x^2-3x-4=0
\\\\
(x-4)(x+1)=0
\\\\
x=\{ -1, 4 \}
.\end{array}
If $z=11$, then,
\begin{array}{l}\require{cancel}
x^2-3x=11
\\\\
x^2-3x-11=0
\\\\
x=\dfrac{3\pm\sqrt{53}}{2}
.\end{array}
The proposed solutions are $
x=\left\{ \dfrac{3\pm\sqrt{53}}{2},-1,4 .\right\}
$ Upon checking, only $
x=\{ -1,4 \}
$ satisfy the original equation.
