$\displaystyle x^{4}+1x^{2}-42=0$

$\displaystyle Let A=x^{2}$

$\displaystyle A^2+A-42=0$

$\displaystyle \left ( A+7 \right )\left ( A-6 \right )=0$

$\displaystyle x^{2}=-7$ $\displaystyle x^{2}=6$

$\displaystyle x=\pm \sqrt{-7} x=\pm \sqrt{6}$

I need help I don't think I did this correctly.