(3w)(w)(w+2)=135

Solution for (3w)(w)(w+2)=135 equation:

(3w)(w)(w+2)=135

We move all terms to the left:

(3w)(w)(w+2)-(135)=0

We multiply parentheses

3w^2+6w-135=0

a = 3; b = 6; c = -135;
Δ = b2-4ac
Δ = 62-4·3·(-135)
Δ = 1656
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:

$\sqrt{\Delta}=\sqrt{1656}=\sqrt{36*46}=\sqrt{36}*\sqrt{46}=6\sqrt{46}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{46}}{2*3}=\frac{-6-6\sqrt{46}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{46}}{2*3}=\frac{-6+6\sqrt{46}}{6}
$