#### Answer

{$0,\pm \sqrt 5$}

#### Work Step by Step

Since, $u^2-u=6$ or, $u^2-u+6=0$
Factorize the expression as follows: $(u-3)(u+2)=0$
or, $u=${$-2,3$}
Replace $u$ with $x^2-2$, we have
$x^2-2=-2 \implies x=0$
and $x^2-2=3 \implies x^2=5$
or, $x=\pm \sqrt 5$
Hence, our solution set is $x=${$0,\pm \sqrt 5$}