3/(z+3)=5-z

Solution for 3/(z+3)=5-z equation:

3/(z+3)=5-z

We move all terms to the left:

3/(z+3)-(5-z)=0


Domain of the equation: (z+3)!=0

We move all terms containing z to the left, all other terms to the right

z!=-3

z∈R


We add all the numbers together, and all the variables

3/(z+3)-(-1z+5)=0

We get rid of parentheses

3/(z+3)+1z-5=0

We multiply all the terms by the denominator


1z*(z+3)-5*(z+3)+3=0

We multiply parentheses

z^2+3z-5z-15+3=0

We add all the numbers together, and all the variables

z^2-2z-12=0

a = 1; b = -2; c = -12;
Δ = b2-4ac
Δ = -22-4·1·(-12)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*1}=\frac{2-2\sqrt{13}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*1}=\frac{2+2\sqrt{13}}{2}
$