8/4z-7=5/9z

Solution for 8/4z-7=5/9z equation:

8/4z-7=5/9z

We move all terms to the left:

8/4z-7-(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

8/4z-(+5/9z)-7=0

We get rid of parentheses

8/4z-5/9z-7=0

We calculate fractions


72z/36z^2+(-20z)/36z^2-7=0

We multiply all the terms by the denominator


72z+(-20z)-7*36z^2=0

Wy multiply elements

-252z^2+72z+(-20z)=0

We get rid of parentheses

-252z^2+72z-20z=0

We add all the numbers together, and all the variables

-252z^2+52z=0

a = -252; b = 52; c = 0;
Δ = b2-4ac
Δ = 522-4·(-252)·0
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-52}{2*-252}=\frac{-104}{-504}
=13/63
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+52}{2*-252}=\frac{0}{-504}
=0
$