3(z-2)-11z=2/3z+2

Solution for 3(z-2)-11z=2/3z+2 equation:

3(z-2)-11z=2/3z+2

We move all terms to the left:

3(z-2)-11z-(2/3z+2)=0


Domain of the equation: 3z+2)!=0

z∈R


We add all the numbers together, and all the variables

-11z+3(z-2)-(2/3z+2)=0

We multiply parentheses

-11z+3z-(2/3z+2)-6=0

We get rid of parentheses

-11z+3z-2/3z-2-6=0

We multiply all the terms by the denominator


-11z*3z+3z*3z-2*3z-6*3z-2=0

Wy multiply elements

-33z^2+9z^2-6z-18z-2=0

We add all the numbers together, and all the variables

-24z^2-24z-2=0

a = -24; b = -24; c = -2;
Δ = b2-4ac
Δ = -242-4·(-24)·(-2)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{6}}{2*-24}=\frac{24-8\sqrt{6}}{-48}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{6}}{2*-24}=\frac{24+8\sqrt{6}}{-48}
$