7/4z-3=5/9z

Solution for 7/4z-3=5/9z equation:

7/4z-3=5/9z

We move all terms to the left:

7/4z-3-(5/9z)=0


Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R



Domain of the equation: 9z)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

7/4z-(+5/9z)-3=0

We get rid of parentheses

7/4z-5/9z-3=0

We calculate fractions


63z/36z^2+(-20z)/36z^2-3=0

We multiply all the terms by the denominator


63z+(-20z)-3*36z^2=0

Wy multiply elements

-108z^2+63z+(-20z)=0

We get rid of parentheses

-108z^2+63z-20z=0

We add all the numbers together, and all the variables

-108z^2+43z=0

a = -108; b = 43; c = 0;
Δ = b2-4ac
Δ = 432-4·(-108)·0
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-43}{2*-108}=\frac{-86}{-216}
=43/108
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+43}{2*-108}=\frac{0}{-216}
=0
$