42=-4+2y^2

Solution for 42=-4+2y^2 equation:

42=-4+2y^2

We move all terms to the left:

42-(-4+2y^2)=0

We get rid of parentheses

-2y^2+4+42=0

We add all the numbers together, and all the variables

-2y^2+46=0

a = -2; b = 0; c = +46;
Δ = b2-4ac
Δ = 02-4·(-2)·46
Δ = 368
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{23}}{2*-2}=\frac{0-4\sqrt{23}}{-4}
=-\frac{4\sqrt{23}}{-4}
=-\frac{\sqrt{23}}{-1}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{23}}{2*-2}=\frac{0+4\sqrt{23}}{-4}
=\frac{4\sqrt{23}}{-4}
=\frac{\sqrt{23}}{-1}
$