0=3w^2+4w-108

Solution for 0=3w^2+4w-108 equation:

0=3w^2+4w-108

We move all terms to the left:

0-(3w^2+4w-108)=0

We add all the numbers together, and all the variables

-(3w^2+4w-108)=0

We get rid of parentheses

-3w^2-4w+108=0

a = -3; b = -4; c = +108;
Δ = b2-4ac
Δ = -42-4·(-3)·108
Δ = 1312
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:

$\sqrt{\Delta}=\sqrt{1312}=\sqrt{16*82}=\sqrt{16}*\sqrt{82}=4\sqrt{82}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{82}}{2*-3}=\frac{4-4\sqrt{82}}{-6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{82}}{2*-3}=\frac{4+4\sqrt{82}}{-6}
$