6j(3j+2)=0

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

6j(3j+2)=0

We multiply parentheses

18j^2+12j=0

a = 18; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·18·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*18}=\frac{-24}{36}
=-2/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*18}=\frac{0}{36}
=0
$