(2z/9)-(3z/2)+(5z/6)=(-4/9z)+3

Solution for (2z/9)-(3z/2)+(5z/6)=(-4/9z)+3 equation:

(2z/9)-(3z/2)+(5z/6)=(-4/9z)+3

We move all terms to the left:

(2z/9)-(3z/2)+(5z/6)-((-4/9z)+3)=0


Domain of the equation: 9z)+3)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+2z/9)-(+3z/2)+(+5z/6)-((-4/9z)+3)=0

We get rid of parentheses

2z/9-3z/2+5z/6-((-4/9z)+3)=0

We calculate fractions


(-8748z^2)/5832z+1620z^2/5832z+144z/5832z+()/5832z=0

We multiply all the terms by the denominator


(-8748z^2)+1620z^2+144z+()=0

We add all the numbers together, and all the variables

1620z^2+(-8748z^2)+144z=0

We get rid of parentheses

1620z^2-8748z^2+144z=0

We add all the numbers together, and all the variables

-7128z^2+144z=0

a = -7128; b = 144; c = 0;
Δ = b2-4ac
Δ = 1442-4·(-7128)·0
Δ = 20736
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{20736}=144$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(144)-144}{2*-7128}=\frac{-288}{-14256}
=2/99
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(144)+144}{2*-7128}=\frac{0}{-14256}
=0
$