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

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

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

We multiply parentheses ..

(+6b^2+3b+4b+2)=0

We get rid of parentheses

6b^2+3b+4b+2=0

We add all the numbers together, and all the variables

6b^2+7b+2=0

a = 6; b = 7; c = +2;
Δ = b2-4ac
Δ = 72-4·6·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*6}=\frac{-8}{12}
=-2/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*6}=\frac{-6}{12}
=-1/2
$