160=10x^2-20x

Solution for 160=10x^2-20x equation:

160=10x^2-20x

We move all terms to the left:

160-(10x^2-20x)=0

We get rid of parentheses

-10x^2+20x+160=0

a = -10; b = 20; c = +160;
Δ = b2-4ac
Δ = 202-4·(-10)·160
Δ = 6800
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:

$\sqrt{\Delta}=\sqrt{6800}=\sqrt{400*17}=\sqrt{400}*\sqrt{17}=20\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{17}}{2*-10}=\frac{-20-20\sqrt{17}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{17}}{2*-10}=\frac{-20+20\sqrt{17}}{-20}
$