1/2j+6=3-2j

Solution for 1/2j+6=3-2j equation:

1/2j+6=3-2j

We move all terms to the left:

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


Domain of the equation: 2j!=0

j!=0/2

j!=0

j∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

4j^2-6j+12j+1=0

We add all the numbers together, and all the variables

4j^2+6j+1=0

a = 4; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·4·1
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{5}}{2*4}=\frac{-6-2\sqrt{5}}{8}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{5}}{2*4}=\frac{-6+2\sqrt{5}}{8}
$