1/z+4=5/2z-8

Solution for 1/z+4=5/2z-8 equation:

1/z+4=5/2z-8

We move all terms to the left:

1/z+4-(5/2z-8)=0


Domain of the equation: z!=0

z∈R



Domain of the equation: 2z-8)!=0

z∈R


We get rid of parentheses

1/z-5/2z+8+4=0

We calculate fractions


2z/2z^2+(-5z)/2z^2+8+4=0

We add all the numbers together, and all the variables

2z/2z^2+(-5z)/2z^2+12=0

We multiply all the terms by the denominator


2z+(-5z)+12*2z^2=0

Wy multiply elements

24z^2+2z+(-5z)=0

We get rid of parentheses

24z^2+2z-5z=0

We add all the numbers together, and all the variables

24z^2-3z=0

a = 24; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·24·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*24}=\frac{0}{48}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*24}=\frac{6}{48}
=1/8
$