0=2x^2+4x-80

Solution for 0=2x^2+4x-80 equation:

0=2x^2+4x-80

We move all terms to the left:

0-(2x^2+4x-80)=0

We add all the numbers together, and all the variables

-(2x^2+4x-80)=0

We get rid of parentheses

-2x^2-4x+80=0

a = -2; b = -4; c = +80;
Δ = b2-4ac
Δ = -42-4·(-2)·80
Δ = 656
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{41}}{2*-2}=\frac{4-4\sqrt{41}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{41}}{2*-2}=\frac{4+4\sqrt{41}}{-4}
$