first use variable y to represent x^2. This means:
$${{\mathtt{Y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{Y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}} = {\mathtt{0}}$$
This is easily solvable by factorization:
$$\left(\left({\mathtt{Y}}{\mathtt{\,-\,}}{\mathtt{2}}\right){\mathtt{\,\times\,}}\left({\mathtt{Y}}{\mathtt{\,-\,}}{\mathtt{4}}\right)\right) = {\mathtt{0}}$$
therefore Y=2 and Y=4 are the answers. Now you can find Xs that satisfy the following:
$${{\mathtt{X}}}^{{\mathtt{2}}} = {\mathtt{4}}$$
and
$${{\mathtt{X}}}^{{\mathtt{2}}} = {\mathtt{2}}$$
.
