(j-3)(2j-1)=0

Solution for (j-3)(2j-1)=0 equation:

(j-3)(2j-1)=0

We multiply parentheses ..

(+2j^2-1j-6j+3)=0

We get rid of parentheses

2j^2-1j-6j+3=0

We add all the numbers together, and all the variables

2j^2-7j+3=0

a = 2; b = -7; c = +3;
Δ = b2-4ac
Δ = -72-4·2·3
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-5}{2*2}=\frac{2}{4}
=1/2
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+5}{2*2}=\frac{12}{4}
=3
$