0=16x^2+160x

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

0=16x^2+160x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-16x^2-160x=0

a = -16; b = -160; c = 0;
Δ = b2-4ac
Δ = -1602-4·(-16)·0
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-160}{2*-16}=\frac{0}{-32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+160}{2*-16}=\frac{320}{-32}
=-10
$