## Elementary Algebra

$x=6$
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.