2+c/3c=1/c

Solution for 2+c/3c=1/c equation:

2+c/3c=1/c

We move all terms to the left:

2+c/3c-(1/c)=0


Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R



Domain of the equation: c)!=0

c!=0/1

c!=0

c∈R


We add all the numbers together, and all the variables

c/3c-(+1/c)+2=0

We get rid of parentheses

c/3c-1/c+2=0

We calculate fractions


c^2/3c^2+(-3c)/3c^2+2=0

We multiply all the terms by the denominator


c^2+(-3c)+2*3c^2=0

Wy multiply elements

c^2+6c^2+(-3c)=0

We get rid of parentheses

c^2+6c^2-3c=0

We add all the numbers together, and all the variables

7c^2-3c=0

a = 7; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·7·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*7}=\frac{0}{14}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*7}=\frac{6}{14}
=3/7
$