5/6z=5)+(-1/3z-4

Solution for 5/6z=5)+(-1/3z-4 equation:

5/6z=5)+(-1/3z-4

We move all terms to the left:

5/6z-(5)+(-1/3z-4)=0


Domain of the equation: 6z!=0

z!=0/6

z!=0

z∈R



Domain of the equation: 3z-4)!=0

z∈R


We get rid of parentheses

5/6z-1/3z-4-5=0

We calculate fractions


15z/18z^2+(-6z)/18z^2-4-5=0

We add all the numbers together, and all the variables

15z/18z^2+(-6z)/18z^2-9=0

We multiply all the terms by the denominator


15z+(-6z)-9*18z^2=0

Wy multiply elements

-162z^2+15z+(-6z)=0

We get rid of parentheses

-162z^2+15z-6z=0

We add all the numbers together, and all the variables

-162z^2+9z=0

a = -162; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-162)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-162}=\frac{-18}{-324}
=1/18
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-162}=\frac{0}{-324}
=0
$