160=20x^2-20x+120

Solution for 160=20x^2-20x+120 equation:

160=20x^2-20x+120

We move all terms to the left:

160-(20x^2-20x+120)=0

We get rid of parentheses

-20x^2+20x-120+160=0

We add all the numbers together, and all the variables

-20x^2+20x+40=0

a = -20; b = 20; c = +40;
Δ = b2-4ac
Δ = 202-4·(-20)·40
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-60}{2*-20}=\frac{-80}{-40}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+60}{2*-20}=\frac{40}{-40}
=-1
$