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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3k^2+8k+4=0

a = 3; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·3·4
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*3}=\frac{-12}{6}
=-2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*3}=\frac{-4}{6}
=-2/3
$