## Elementary Algebra

$n=3$
Squaring both sides of the equation and using the properties of radicals, the solution to the given equation is \begin{array}{l}\require{cancel}\left( \sqrt{n-3} \right)^2=\left( 3-n \right)^2 \\\\ n-3=(3)^2+2(3)(-n)+(-n)^2 \\\\ n-3=9-6n+n^2 \\\\ 0=(9+3)+(-6n-n)+n^2 \\\\ 12-7n+n^2=0 \\\\ (4-n)(3-n)=0 \\\\ n=\{3,4\} .\end{array} Upon checking, only $n=3$ satisfies the original equation.