What are the zeros of y-7=4x^2
and Solve square root of 2x^2 - sqaure root of 2 = 0
to [1].
If I understand this question correctly then y = 0. Therefore solve for x:
$\displaystyle -7 = 4x^2~\implies~x^2=-\dfrac74~\implies~x=\pm i \cdot \dfrac12 \sqrt{7}$
to [2]:
$\displaystyle \sqrt{2x^2} - \sqrt{2} =0~\implies~\sqrt{2} \cdot \sqrt{x^2} = \sqrt{2} \implies~|x| = 1$