3j(j-2)=2(j-2)

Solution for 3j(j-2)=2(j-2) equation:

3j(j-2)=2(j-2)

We move all terms to the left:

3j(j-2)-(2(j-2))=0

We multiply parentheses

3j^2-6j-(2(j-2))=0


We calculate terms in parentheses: -(2(j-2)), so:

2(j-2)

We multiply parentheses

2j-4

Back to the equation:

-(2j-4)


We get rid of parentheses

3j^2-6j-2j+4=0

We add all the numbers together, and all the variables

3j^2-8j+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:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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