In this equation we can't complete the square (cause there's no linear term) but we can solve it, so let's do that:
[tex]\begin{gathered} 6x^2-6=90 \\ 6x^2=96 \\ x^2=\frac{96}{6} \\ x^2=16 \\ x=\pm\sqrt[]{16} \\ x=\pm4 \end{gathered}[/tex]Therefore x=4 or x=-4.
