8/3z+11/9z-7/3=7/6

Solution for 8/3z+11/9z-7/3=7/6 equation:

8/3z+11/9z-7/3=7/6

We move all terms to the left:

8/3z+11/9z-7/3-(7/6)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R



Domain of the equation: 9z!=0

z!=0/9

z!=0

z∈R


We add all the numbers together, and all the variables

8/3z+11/9z-7/3-(+7/6)=0

We get rid of parentheses

8/3z+11/9z-7/3-7/6=0

We calculate fractions


(-5103z^2)/1458z^2+2592z/1458z^2+1782z/1458z^2+(-2268z)/1458z^2=0

We multiply all the terms by the denominator


(-5103z^2)+2592z+1782z+(-2268z)=0

We add all the numbers together, and all the variables

(-5103z^2)+4374z+(-2268z)=0

We get rid of parentheses

-5103z^2+4374z-2268z=0

We add all the numbers together, and all the variables

-5103z^2+2106z=0

a = -5103; b = 2106; c = 0;
Δ = b2-4ac
Δ = 21062-4·(-5103)·0
Δ = 4435236
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4435236}=2106$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2106)-2106}{2*-5103}=\frac{-4212}{-10206}
=26/63
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2106)+2106}{2*-5103}=\frac{0}{-10206}
=0
$