160=(4+2w)w

Solution for 160=(4+2w)w equation:

160=(4+2w)w

We move all terms to the left:

160-((4+2w)w)=0

We add all the numbers together, and all the variables

-((2w+4)w)+160=0


We calculate terms in parentheses: -((2w+4)w), so:

(2w+4)w

We multiply parentheses

2w^2+4w

Back to the equation:

-(2w^2+4w)


We get rid of parentheses

-2w^2-4w+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:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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