(3w)(w)(2+2w)=216

Solution for (3w)(w)(2+2w)=216 equation:

(3w)(w)(2+2w)=216

We move all terms to the left:

(3w)(w)(2+2w)-(216)=0

We add all the numbers together, and all the variables

3ww(2w+2)-216=0

We multiply parentheses

6w^2+6w-216=0

a = 6; b = 6; c = -216;
Δ = b2-4ac
Δ = 62-4·6·(-216)
Δ = 5220
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:

$\sqrt{\Delta}=\sqrt{5220}=\sqrt{36*145}=\sqrt{36}*\sqrt{145}=6\sqrt{145}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{145}}{2*6}=\frac{-6-6\sqrt{145}}{12}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{145}}{2*6}=\frac{-6+6\sqrt{145}}{12}
$