z/6z+3-4z=9

Solution for z/6z+3-4z=9 equation:

z/6z+3-4z=9

We move all terms to the left:

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


Domain of the equation: 6z!=0

z!=0/6

z!=0

z∈R


We add all the numbers together, and all the variables

-4z+z/6z-6=0

We multiply all the terms by the denominator


-4z*6z+z-6*6z=0

We add all the numbers together, and all the variables

z-4z*6z-6*6z=0

Wy multiply elements

-24z^2+z-36z=0

We add all the numbers together, and all the variables

-24z^2-35z=0

a = -24; b = -35; c = 0;
Δ = b2-4ac
Δ = -352-4·(-24)·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-35}{2*-24}=\frac{0}{-48}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+35}{2*-24}=\frac{70}{-48}
=-1+11/24
$