2/3z+32=z+6

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

2/3z+32=z+6

We move all terms to the left:

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


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


-z*3z-6*3z+32*3z+2=0

Wy multiply elements

-3z^2-18z+96z+2=0

We add all the numbers together, and all the variables

-3z^2+78z+2=0

a = -3; b = 78; c = +2;
Δ = b2-4ac
Δ = 782-4·(-3)·2
Δ = 6108
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:

$\sqrt{\Delta}=\sqrt{6108}=\sqrt{4*1527}=\sqrt{4}*\sqrt{1527}=2\sqrt{1527}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(78)-2\sqrt{1527}}{2*-3}=\frac{-78-2\sqrt{1527}}{-6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(78)+2\sqrt{1527}}{2*-3}=\frac{-78+2\sqrt{1527}}{-6}
$