-1/z+4/3z=2/z-5

Solution for -1/z+4/3z=2/z-5 equation:

-1/z+4/3z=2/z-5

We move all terms to the left:

-1/z+4/3z-(2/z-5)=0


Domain of the equation: z!=0

z∈R



Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R



Domain of the equation: z-5)!=0

z∈R


We get rid of parentheses

-1/z+4/3z-2/z+5=0

We calculate fractions


(-6z-1)/3z^2+4z/3z^2+5=0

We multiply all the terms by the denominator


(-6z-1)+4z+5*3z^2=0

We add all the numbers together, and all the variables

4z+(-6z-1)+5*3z^2=0

Wy multiply elements

15z^2+4z+(-6z-1)=0

We get rid of parentheses

15z^2+4z-6z-1=0

We add all the numbers together, and all the variables

15z^2-2z-1=0

a = 15; b = -2; c = -1;
Δ = b2-4ac
Δ = -22-4·15·(-1)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*15}=\frac{-6}{30}
=-1/5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*15}=\frac{10}{30}
=1/3
$