z+(8/z)=-9

Solution for z+(8/z)=-9 equation:

z+(8/z)=-9

We move all terms to the left:

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


Domain of the equation: z)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

z+(+8/z)+9=0

We get rid of parentheses

z+8/z+9=0

We multiply all the terms by the denominator


z*z+9*z+8=0

We add all the numbers together, and all the variables

9z+z*z+8=0

Wy multiply elements

z^2+9z+8=0

a = 1; b = 9; c = +8;
Δ = b2-4ac
Δ = 92-4·1·8
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-7}{2*1}=\frac{-16}{2}
=-8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+7}{2*1}=\frac{-2}{2}
=-1
$