Perfect Squares

Let, $x+168=a^2$ and $x+100=b^2$ .Therefore $a^2-b^2=68$ .We divided into three cases. $a+b=68,a-b=1$ no solution in this case. $a+b=34,a-b=2 \implies a=18,b=16$ $a+b=17,a-b=4$ no solution in this case. Therefore after evaluating we get only on solution that $\boxed{x=156}$