1/36z+18=3z-9

Solution for 1/36z+18=3z-9 equation:

1/36z+18=3z-9

We move all terms to the left:

1/36z+18-(3z-9)=0


Domain of the equation: 36z!=0

z!=0/36

z!=0

z∈R


We get rid of parentheses

1/36z-3z+9+18=0

We multiply all the terms by the denominator


-3z*36z+9*36z+18*36z+1=0

Wy multiply elements

-108z^2+324z+648z+1=0

We add all the numbers together, and all the variables

-108z^2+972z+1=0

a = -108; b = 972; c = +1;
Δ = b2-4ac
Δ = 9722-4·(-108)·1
Δ = 945216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{945216}=\sqrt{576*1641}=\sqrt{576}*\sqrt{1641}=24\sqrt{1641}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(972)-24\sqrt{1641}}{2*-108}=\frac{-972-24\sqrt{1641}}{-216}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(972)+24\sqrt{1641}}{2*-108}=\frac{-972+24\sqrt{1641}}{-216}
$