160=2x^2+4x

Solution for 160=2x^2+4x equation:

160=2x^2+4x

We move all terms to the left:

160-(2x^2+4x)=0

We get rid of parentheses

-2x^2-4x+160=0

a = -2; b = -4; c = +160;
Δ = b2-4ac
Δ = -42-4·(-2)·160
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-36}{2*-2}=\frac{-32}{-4}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+36}{2*-2}=\frac{40}{-4}
=-10
$