0=-16x^2+32x+128

Solution for 0=-16x^2+32x+128 equation:

0=-16x^2+32x+128

We move all terms to the left:

0-(-16x^2+32x+128)=0

We add all the numbers together, and all the variables

-(-16x^2+32x+128)=0

We get rid of parentheses

16x^2-32x-128=0

a = 16; b = -32; c = -128;
Δ = b2-4ac
Δ = -322-4·16·(-128)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-96}{2*16}=\frac{-64}{32}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+96}{2*16}=\frac{128}{32}
=4
$