Z^2+3z-4/z+4=z-1

Solution for Z^2+3z-4/z+4=z-1 equation:

^2+3Z-4/Z+4=Z-1

We move all terms to the left:

^2+3Z-4/Z+4-(Z-1)=0


Domain of the equation: Z!=0

Z∈R


We add all the numbers together, and all the variables

3Z-4/Z-(Z-1)=0

We get rid of parentheses

3Z-4/Z-Z+1=0

We multiply all the terms by the denominator


3Z*Z-Z*Z+1*Z-4=0

We add all the numbers together, and all the variables

Z+3Z*Z-Z*Z-4=0

Wy multiply elements

3Z^2-1Z^2+Z-4=0

We add all the numbers together, and all the variables

2Z^2+Z-4=0

a = 2; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·2·(-4)
Δ = 33
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{33}}{2*2}=\frac{-1-\sqrt{33}}{4}
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{33}}{2*2}=\frac{-1+\sqrt{33}}{4}
$