2y^2-20-48=0

Solution for 2y^2-20-48=0 equation:

2y^2-20-48=0

We add all the numbers together, and all the variables

2y^2-68=0

a = 2; b = 0; c = -68;
Δ = b2-4ac
Δ = 02-4·2·(-68)
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{34}}{2*2}=\frac{0-4\sqrt{34}}{4}
=-\frac{4\sqrt{34}}{4}
=-\sqrt{34}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{34}}{2*2}=\frac{0+4\sqrt{34}}{4}
=\frac{4\sqrt{34}}{4}
=\sqrt{34}
$