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

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

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

We move all terms to the left:

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


Domain of the equation: j!=0

j∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

-2/j-j+3=0

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

3j-j*j-2=0

Wy multiply elements

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

a = -1; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·(-1)·(-2)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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