(4/5)z=2/9

Solution for (4/5)z=2/9 equation:

(4/5)z=2/9

We move all terms to the left:

(4/5)z-(2/9)=0


Domain of the equation: 5)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+4/5)z-(+2/9)=0

We multiply parentheses

4z^2-(+2/9)=0

We get rid of parentheses

4z^2-2/9=0

We multiply all the terms by the denominator


4z^2*9-2=0

Wy multiply elements

36z^2-2=0

a = 36; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·36·(-2)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*36}=\frac{0-12\sqrt{2}}{72}
=-\frac{12\sqrt{2}}{72}
=-\frac{\sqrt{2}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*36}=\frac{0+12\sqrt{2}}{72}
=\frac{12\sqrt{2}}{72}
=\frac{\sqrt{2}}{6}
$