0=-16x^2+32x+256

Solution for 0=-16x^2+32x+256 equation:

0=-16x^2+32x+256

We move all terms to the left:

0-(-16x^2+32x+256)=0

We add all the numbers together, and all the variables

-(-16x^2+32x+256)=0

We get rid of parentheses

16x^2-32x-256=0

a = 16; b = -32; c = -256;
Δ = b2-4ac
Δ = -322-4·16·(-256)
Δ = 17408
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:

$\sqrt{\Delta}=\sqrt{17408}=\sqrt{1024*17}=\sqrt{1024}*\sqrt{17}=32\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-32\sqrt{17}}{2*16}=\frac{32-32\sqrt{17}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+32\sqrt{17}}{2*16}=\frac{32+32\sqrt{17}}{32}
$