(7/3z)+(6/z)=-4

Solution for (7/3z)+(6/z)=-4 equation:

(7/3z)+(6/z)=-4

We move all terms to the left:

(7/3z)+(6/z)-(-4)=0


Domain of the equation: 3z)!=0

z!=0/1

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

(+7/3z)+(+6/z)-(-4)=0

We add all the numbers together, and all the variables

(+7/3z)+(+6/z)+4=0

We get rid of parentheses

7/3z+6/z+4=0

We calculate fractions


7z/3z^2+18z/3z^2+4=0

We multiply all the terms by the denominator


7z+18z+4*3z^2=0

We add all the numbers together, and all the variables

25z+4*3z^2=0

Wy multiply elements

12z^2+25z=0

a = 12; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·12·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*12}=\frac{-50}{24}
=-2+1/12
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*12}=\frac{0}{24}
=0
$