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

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

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

We move all terms to the left:

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


Domain of the equation: z!=0

z∈R



Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R



Domain of the equation: z-6)!=0

z∈R


We get rid of parentheses

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

We calculate fractions


(-6z-2)/2z^2+z/2z^2+6=0

We multiply all the terms by the denominator


(-6z-2)+z+6*2z^2=0

We add all the numbers together, and all the variables

z+(-6z-2)+6*2z^2=0

Wy multiply elements

12z^2+z+(-6z-2)=0

We get rid of parentheses

12z^2+z-6z-2=0

We add all the numbers together, and all the variables

12z^2-5z-2=0

a = 12; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·12·(-2)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*12}=\frac{-6}{24}
=-1/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*12}=\frac{16}{24}
=2/3
$