-16x^2+160x+128=0

Solution for -16x^2+160x+128=0 equation:

-16x^2+160x+128=0

a = -16; b = 160; c = +128;
Δ = b2-4ac
Δ = 1602-4·(-16)·128
Δ = 33792
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:

$\sqrt{\Delta}=\sqrt{33792}=\sqrt{1024*33}=\sqrt{1024}*\sqrt{33}=32\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-32\sqrt{33}}{2*-16}=\frac{-160-32\sqrt{33}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+32\sqrt{33}}{2*-16}=\frac{-160+32\sqrt{33}}{-32}
$