320=2y^2

Solution for 320=2y^2 equation:

320=2y^2

We move all terms to the left:

320-(2y^2)=0

a = -2; b = 0; c = +320;
Δ = b2-4ac
Δ = 02-4·(-2)·320
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*-2}=\frac{0-16\sqrt{10}}{-4}
=-\frac{16\sqrt{10}}{-4}
=-\frac{4\sqrt{10}}{-1}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*-2}=\frac{0+16\sqrt{10}}{-4}
=\frac{16\sqrt{10}}{-4}
=\frac{4\sqrt{10}}{-1}
$