Answer
$x=6$
Work Step by Step
Squaring both sides of the equation and then using the properties of equality, we obtain:
\begin{array}{l}\require{cancel}\left(
\sqrt{-2x+28}
\right)^2=\left(
x-2
\right)^2
\\\\
-2x+28=(x)^2+2(x)(-2)+(-2)^2
\\\\
-2x+28=x^2-4x+4
\\\\
0=x^2+(-4x+2x)+(4-28)
\\\\
x^2-2x-24=0
\\\\
(x-6)(x+4)=0
\\\\
x=\{-4,6\}
.\end{array}
Upon checking, only $
x=6
$ satisfies the original equation.
