(3/5)z+1=25/9

Solution for (3/5)z+1=25/9 equation:

(3/5)z+1=25/9

We move all terms to the left:

(3/5)z+1-(25/9)=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

(+3/5)z+1-(+25/9)=0

We multiply parentheses

3z^2+1-(+25/9)=0

We get rid of parentheses

3z^2+1-25/9=0

We multiply all the terms by the denominator


3z^2*9-25+1*9=0

We add all the numbers together, and all the variables

3z^2*9-16=0

Wy multiply elements

27z^2-16=0

a = 27; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·27·(-16)
Δ = 1728
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:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{3}}{2*27}=\frac{0-24\sqrt{3}}{54}
=-\frac{24\sqrt{3}}{54}
=-\frac{4\sqrt{3}}{9}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{3}}{2*27}=\frac{0+24\sqrt{3}}{54}
=\frac{24\sqrt{3}}{54}
=\frac{4\sqrt{3}}{9}
$