(4w+12)w=160

Solution for (4w+12)w=160 equation:

(4w+12)w=160

We move all terms to the left:

(4w+12)w-(160)=0

We multiply parentheses

4w^2+12w-160=0

a = 4; b = 12; c = -160;
Δ = b2-4ac
Δ = 122-4·4·(-160)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-52}{2*4}=\frac{-64}{8}
=-8
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+52}{2*4}=\frac{40}{8}
=5
$