#### 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{x^2-35}
\right)^2=\left(
x-5
\right)^2
\\\\
x^2-35=(x)^2+2(x)(-5)+(-5)^2
\\\\
x^2-35=x^2-10x+25
\\\\
(x^2-x^2)+10x=25+35
\\\\
10x=60
\\\\
x=\dfrac{60}{10}
\\\\
x=6
.\end{array}
Upon checking, $
x=6
$ satisfies the original equation.