2/3z=9+3/4z=

Solution for 2/3z=9+3/4z= equation:

2/3z=9+3/4z=

We move all terms to the left:

2/3z-(9+3/4z)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R



Domain of the equation: 4z)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

2/3z-(3/4z+9)=0

We get rid of parentheses

2/3z-3/4z-9=0

We calculate fractions


8z/12z^2+(-9z)/12z^2-9=0

We multiply all the terms by the denominator


8z+(-9z)-9*12z^2=0

Wy multiply elements

-108z^2+8z+(-9z)=0

We get rid of parentheses

-108z^2+8z-9z=0

We add all the numbers together, and all the variables

-108z^2-1z=0

a = -108; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-108)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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