(3w+2)(1+w)=0

Solution for (3w+2)(1+w)=0 equation:

(3w+2)(1+w)=0

We add all the numbers together, and all the variables

(3w+2)(w+1)=0

We multiply parentheses ..

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

We get rid of parentheses

3w^2+3w+2w+2=0

We add all the numbers together, and all the variables

3w^2+5w+2=0

a = 3; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·3·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*3}=\frac{-6}{6}
=-1
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*3}=\frac{-4}{6}
=-2/3
$