0=16x^2+80x-128

Solution for 0=16x^2+80x-128 equation:

0=16x^2+80x-128

We move all terms to the left:

0-(16x^2+80x-128)=0

We add all the numbers together, and all the variables

-(16x^2+80x-128)=0

We get rid of parentheses

-16x^2-80x+128=0

a = -16; b = -80; c = +128;
Δ = b2-4ac
Δ = -802-4·(-16)·128
Δ = 14592
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:

$\sqrt{\Delta}=\sqrt{14592}=\sqrt{256*57}=\sqrt{256}*\sqrt{57}=16\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-16\sqrt{57}}{2*-16}=\frac{80-16\sqrt{57}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+16\sqrt{57}}{2*-16}=\frac{80+16\sqrt{57}}{-32}
$