Answer
$x=\left\{ -1, 2 .\right\}$
Work Step by Step
Let $z=
x^2-x
.$ Then the given equation, $
\sqrt{(x^2-x)+7}=2(x^2-x)-1
$, is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{z+7}=2(z)-1
\\\\
\sqrt{z+7}=2z-1
.\end{array}
Squaring both sides of the equal sign results to
\begin{array}{l}\require{cancel}
z+7=(2z)^2+2(2z)(-1)+(-1)^2
\\\\
z+7=4z^2-4z+1
\\\\
-4z^2+(z+4z)+(7-1)=0
\\\\
-4z^2+5z+6=0
\\\\
4z^2-5z-6=0
\\\\
(4z+3)(z-2)=0
\\\\
z=\left\{ -\dfrac{3}{4}, 2\right\}
.\end{array}
If $z=-\dfrac{3}{4}$, then,
\begin{array}{l}\require{cancel}
x^2-x=-\dfrac{3}{4}
\\\\
4(x^2-x)=-3
\\\\
4x^2-4x=-3
\\\\
4x^2-4x+3=0
\\\\
x=\text{ not a real number}
.\end{array}
If $z=2$, then,
\begin{array}{l}\require{cancel}
x^2-x=2
\\\\
x^2-x-2=0
\\\\
(x-2)(x+1)=0
\\\\
x=\{ -1,2 \}
.\end{array}
The proposed solutions are $
x=\left\{ -1, 2 .\right\}
$ Upon checking, all the proposed solutions satisfy the original equation. Hence, $
x=\left\{ -1, 2 .\right\}
.$
