7/2z+6=3/z

Solution for 7/2z+6=3/z equation:

7/2z+6=3/z

We move all terms to the left:

7/2z+6-(3/z)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R



Domain of the equation: z)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

7/2z-(+3/z)+6=0

We get rid of parentheses

7/2z-3/z+6=0

We calculate fractions


7z/2z^2+(-6z)/2z^2+6=0

We multiply all the terms by the denominator


7z+(-6z)+6*2z^2=0

Wy multiply elements

12z^2+7z+(-6z)=0

We get rid of parentheses

12z^2+7z-6z=0

We add all the numbers together, and all the variables

12z^2+z=0

a = 12; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·12·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*12}=\frac{-2}{24}
=-1/12
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*12}=\frac{0}{24}
=0
$