-16x^2+96x+0=128

Solution for -16x^2+96x+0=128 equation:

-16x^2+96x+0=128

We move all terms to the left:

-16x^2+96x+0-(128)=0

We add all the numbers together, and all the variables

-16x^2+96x-128=0

a = -16; b = 96; c = -128;
Δ = b2-4ac
Δ = 962-4·(-16)·(-128)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-32}{2*-16}=\frac{-128}{-32}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+32}{2*-16}=\frac{-64}{-32}
=+2
$