(4/9)z=12

Solution for (4/9)z=12 equation:

(4/9)z=12

We move all terms to the left:

(4/9)z-(12)=0


Domain of the equation: 9)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+4/9)z-12=0

We multiply parentheses

4z^2-12=0

a = 4; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·4·(-12)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*4}=\frac{0-8\sqrt{3}}{8}
=-\frac{8\sqrt{3}}{8}
=-\sqrt{3}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*4}=\frac{0+8\sqrt{3}}{8}
=\frac{8\sqrt{3}}{8}
=\sqrt{3}
$