(2/3)z=12

Solution for (2/3)z=12 equation:

(2/3)z=12

We move all terms to the left:

(2/3)z-(12)=0


Domain of the equation: 3)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+2/3)z-12=0

We multiply parentheses

2z^2-12=0

a = 2; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·2·(-12)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*2}=\frac{0-4\sqrt{6}}{4}
=-\frac{4\sqrt{6}}{4}
=-\sqrt{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*2}=\frac{0+4\sqrt{6}}{4}
=\frac{4\sqrt{6}}{4}
=\sqrt{6}
$