Answer
$n=3$
Work Step by Step
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.
