(3-z^2)/z=3

Solution for (3-z^2)/z=3 equation:

(3-z^2)/z=3

We move all terms to the left:

(3-z^2)/z-(3)=0


Domain of the equation: z!=0

z∈R


We multiply all the terms by the denominator


(3-z^2)-3*z=0

We add all the numbers together, and all the variables

(3-z^2)-3z=0

We get rid of parentheses

-z^2-3z+3=0

We add all the numbers together, and all the variables

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

a = -1; b = -3; c = +3;
Δ = b2-4ac
Δ = -32-4·(-1)·3
Δ = 21
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{-(-3)-\sqrt{21}}{2*-1}=\frac{3-\sqrt{21}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{21}}{2*-1}=\frac{3+\sqrt{21}}{-2}
$