(2w+4)w=160^2

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

(2w+4)w=160^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

(2w+4)w-25600=0

We multiply parentheses

2w^2+4w-25600=0

a = 2; b = 4; c = -25600;
Δ = b2-4ac
Δ = 42-4·2·(-25600)
Δ = 204816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{204816}=\sqrt{16*12801}=\sqrt{16}*\sqrt{12801}=4\sqrt{12801}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{12801}}{2*2}=\frac{-4-4\sqrt{12801}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{12801}}{2*2}=\frac{-4+4\sqrt{12801}}{4}
$