z=2/3z+32+6

Solution for z=2/3z+32+6 equation:

z=2/3z+32+6

We move all terms to the left:

z-(2/3z+32+6)=0


Domain of the equation: 3z+32+6)!=0

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

3z+6)!=-32

z∈R


We add all the numbers together, and all the variables

z-(2/3z+38)=0

We get rid of parentheses

z-2/3z-38=0

We multiply all the terms by the denominator


z*3z-38*3z-2=0

Wy multiply elements

3z^2-114z-2=0

a = 3; b = -114; c = -2;
Δ = b2-4ac
Δ = -1142-4·3·(-2)
Δ = 13020
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:

$\sqrt{\Delta}=\sqrt{13020}=\sqrt{4*3255}=\sqrt{4}*\sqrt{3255}=2\sqrt{3255}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-114)-2\sqrt{3255}}{2*3}=\frac{114-2\sqrt{3255}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-114)+2\sqrt{3255}}{2*3}=\frac{114+2\sqrt{3255}}{6}
$