j-6=1/2j=3

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

j-6=1/2j=3

We move all terms to the left:

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


Domain of the equation: 2j)!=0

j!=0/1

j!=0

j∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

j-1/2j-6=0

We multiply all the terms by the denominator


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

Wy multiply elements

2j^2-12j-1=0

a = 2; b = -12; c = -1;
Δ = b2-4ac
Δ = -122-4·2·(-1)
Δ = 152
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:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{38}}{2*2}=\frac{12-2\sqrt{38}}{4}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{38}}{2*2}=\frac{12+2\sqrt{38}}{4}
$