2y^2+8=80

Solution for 2y^2+8=80 equation:

2y^2+8=80

We move all terms to the left:

2y^2+8-(80)=0

We add all the numbers together, and all the variables

2y^2-72=0

a = 2; b = 0; c = -72;
Δ = b2-4ac
Δ = 02-4·2·(-72)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*2}=\frac{-24}{4}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*2}=\frac{24}{4}
=6
$