0=-3x^2+160x

Solution for 0=-3x^2+160x equation:

0=-3x^2+160x

We move all terms to the left:

0-(-3x^2+160x)=0

We add all the numbers together, and all the variables

-(-3x^2+160x)=0

We get rid of parentheses

3x^2-160x=0

a = 3; b = -160; c = 0;
Δ = b2-4ac
Δ = -1602-4·3·0
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*3}=\frac{0}{6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+160}{2*3}=\frac{320}{6}
=53+1/3
$