(3k+2)(2k+3)=0

Solution for (3k+2)(2k+3)=0 equation:

(3k+2)(2k+3)=0

We multiply parentheses ..

(+6k^2+9k+4k+6)=0

We get rid of parentheses

6k^2+9k+4k+6=0

We add all the numbers together, and all the variables

6k^2+13k+6=0

a = 6; b = 13; c = +6;
Δ = b2-4ac
Δ = 132-4·6·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*6}=\frac{-18}{12}
=-1+1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*6}=\frac{-8}{12}
=-2/3
$