(3w+2)(6+w)=0

Solution for (3w+2)(6+w)=0 equation:

(3w+2)(6+w)=0

We add all the numbers together, and all the variables

(3w+2)(w+6)=0

We multiply parentheses ..

(+3w^2+18w+2w+12)=0

We get rid of parentheses

3w^2+18w+2w+12=0

We add all the numbers together, and all the variables

3w^2+20w+12=0

a = 3; b = 20; c = +12;
Δ = b2-4ac
Δ = 202-4·3·12
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*3}=\frac{-36}{6}
=-6
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*3}=\frac{-4}{6}
=-2/3
$