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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -((-16x^2)+160x+128), so:

(-16x^2)+160x+128

We get rid of parentheses

-16x^2+160x+128

Back to the equation:

-(-16x^2+160x+128)


We get rid of parentheses

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}
$