Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5 - Page 424: 36

$x=6$

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.