z+3=9/z-6=-2

Solution for z+3=9/z-6=-2 equation:

z+3=9/z-6=-2

We move all terms to the left:

z+3-(9/z-6)=0


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

z∈R


We get rid of parentheses

z-9/z+6+3=0

We multiply all the terms by the denominator


z*z+6*z+3*z-9=0

We add all the numbers together, and all the variables

9z+z*z-9=0

Wy multiply elements

z^2+9z-9=0

a = 1; b = 9; c = -9;
Δ = b2-4ac
Δ = 92-4·1·(-9)
Δ = 117
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{13}}{2*1}=\frac{-9-3\sqrt{13}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{13}}{2*1}=\frac{-9+3\sqrt{13}}{2}
$