8/5z+9/z=1

Solution for 8/5z+9/z=1 equation:

8/5z+9/z=1

We move all terms to the left:

8/5z+9/z-(1)=0


Domain of the equation: 5z!=0

z!=0/5

z!=0

z∈R



Domain of the equation: z!=0

z∈R


We calculate fractions


8z/5z^2+45z/5z^2-1=0

We multiply all the terms by the denominator


8z+45z-1*5z^2=0

We add all the numbers together, and all the variables

53z-1*5z^2=0

Wy multiply elements

-5z^2+53z=0

a = -5; b = 53; c = 0;
Δ = b2-4ac
Δ = 532-4·(-5)·0
Δ = 2809
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2809}=53$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-53}{2*-5}=\frac{-106}{-10}
=10+3/5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+53}{2*-5}=\frac{0}{-10}
=0
$