160=w(2w+3)

Solution for 160=w(2w+3) equation:

160=w(2w+3)

We move all terms to the left:

160-(w(2w+3))=0


We calculate terms in parentheses: -(w(2w+3)), so:

w(2w+3)

We multiply parentheses

2w^2+3w

Back to the equation:

-(2w^2+3w)


We get rid of parentheses

-2w^2-3w+160=0

a = -2; b = -3; c = +160;
Δ = b2-4ac
Δ = -32-4·(-2)·160
Δ = 1289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{1289}}{2*-2}=\frac{3-\sqrt{1289}}{-4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1289}}{2*-2}=\frac{3+\sqrt{1289}}{-4}
$