(3/z-2)-(9/2z+7)+(4/z-2)=0

Solution for (3/z-2)-(9/2z+7)+(4/z-2)=0 equation:

(3/z-2)-(9/2z+7)+(4/z-2)=0


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

z∈R



Domain of the equation: 2z+7)!=0

z∈R


We get rid of parentheses

3/z-9/2z+4/z-2-7-2=0

We calculate fractions


(8z+3)/2z^2+(-9z)/2z^2-2-7-2=0

We add all the numbers together, and all the variables

(8z+3)/2z^2+(-9z)/2z^2-11=0

We multiply all the terms by the denominator


(8z+3)+(-9z)-11*2z^2=0

Wy multiply elements

-22z^2+(8z+3)+(-9z)=0

We get rid of parentheses

-22z^2+8z-9z+3=0

We add all the numbers together, and all the variables

-22z^2-1z+3=0

a = -22; b = -1; c = +3;
Δ = b2-4ac
Δ = -12-4·(-22)·3
Δ = 265
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{265}}{2*-22}=\frac{1-\sqrt{265}}{-44}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{265}}{2*-22}=\frac{1+\sqrt{265}}{-44}
$