0=16x^2-160x+256

Solution for 0=16x^2-160x+256 equation:

0=16x^2-160x+256

We move all terms to the left:

0-(16x^2-160x+256)=0

We add all the numbers together, and all the variables

-(16x^2-160x+256)=0

We get rid of parentheses

-16x^2+160x-256=0

a = -16; b = 160; c = -256;
Δ = b2-4ac
Δ = 1602-4·(-16)·(-256)
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{-(160)-96}{2*-16}=\frac{-256}{-32}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+96}{2*-16}=\frac{-64}{-32}
=+2
$