6j(9j+1)=0

Solution for 6j(9j+1)=0 equation:

6j(9j+1)=0

We multiply parentheses

54j^2+6j=0

a = 54; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·54·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*54}=\frac{-12}{108}
=-1/9
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*54}=\frac{0}{108}
=0
$