7/3z-6z=3

Solution for 7/3z-6z=3 equation:

7/3z-6z=3

We move all terms to the left:

7/3z-6z-(3)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We add all the numbers together, and all the variables

-6z+7/3z-3=0

We multiply all the terms by the denominator


-6z*3z-3*3z+7=0

Wy multiply elements

-18z^2-9z+7=0

a = -18; b = -9; c = +7;
Δ = b2-4ac
Δ = -92-4·(-18)·7
Δ = 585
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:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{65}}{2*-18}=\frac{9-3\sqrt{65}}{-36}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{65}}{2*-18}=\frac{9+3\sqrt{65}}{-36}
$