z=5/3z+4

Solution for z=5/3z+4 equation:

z=5/3z+4

We move all terms to the left:

z-(5/3z+4)=0


Domain of the equation: 3z+4)!=0

z∈R


We get rid of parentheses

z-5/3z-4=0

We multiply all the terms by the denominator


z*3z-4*3z-5=0

Wy multiply elements

3z^2-12z-5=0

a = 3; b = -12; c = -5;
Δ = b2-4ac
Δ = -122-4·3·(-5)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{51}}{2*3}=\frac{12-2\sqrt{51}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{51}}{2*3}=\frac{12+2\sqrt{51}}{6}
$