0=-40x-2x^2+12800

Solution for 0=-40x-2x^2+12800 equation:

0=-40x-2x^2+12800

We move all terms to the left:

0-(-40x-2x^2+12800)=0

We add all the numbers together, and all the variables

-(-40x-2x^2+12800)=0

We get rid of parentheses

2x^2+40x-12800=0

a = 2; b = 40; c = -12800;
Δ = b2-4ac
Δ = 402-4·2·(-12800)
Δ = 104000
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:

$\sqrt{\Delta}=\sqrt{104000}=\sqrt{1600*65}=\sqrt{1600}*\sqrt{65}=40\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{65}}{2*2}=\frac{-40-40\sqrt{65}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{65}}{2*2}=\frac{-40+40\sqrt{65}}{4}
$