(3/c+2)-(2/2c-3)=1/7

Solution for (3/c+2)-(2/2c-3)=1/7 equation:

3/c+2-2/2c-3=1/7

We move all terms to the left:

3/c+2-2/2c-3-(1/7)=0


Domain of the equation: c!=0

c∈R



Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R


We add all the numbers together, and all the variables

3/c-2/2c+2-3-(+1/7)=0

We add all the numbers together, and all the variables

3/c-2/2c-1-(+1/7)=0

We get rid of parentheses

3/c-2/2c-1-1/7=0

We calculate fractions


(-4c^2)/98c^2+294c/98c^2+(-98c)/98c^2-1=0

We multiply all the terms by the denominator


(-4c^2)+294c+(-98c)-1*98c^2=0

Wy multiply elements

(-4c^2)-98c^2+294c+(-98c)=0

We get rid of parentheses

-4c^2-98c^2+294c-98c=0

We add all the numbers together, and all the variables

-102c^2+196c=0

a = -102; b = 196; c = 0;
Δ = b2-4ac
Δ = 1962-4·(-102)·0
Δ = 38416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{38416}=196$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(196)-196}{2*-102}=\frac{-392}{-204}
=1+47/51
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(196)+196}{2*-102}=\frac{0}{-204}
=0
$