just a question
or
Then don't worry about "common denominator" or adding fractions: multiply the entire equation by (3x-2c)(2x- 3c)(2c) (which is, actually, the "common denominator.)
Multiplying the first term by that, , the "3x- 2c" terms cancel leaving . Multiplying the second term by that, , the "2x-3c" terms cancel leaving . Finally, multiplying the right hand side of the equation by that, , the "2c" terms cancel leaving .
Putting those together, multiplying the entire equation by (3x-2c)(2x-3c)(2c) changes the equation to or , . Dividing through by 2, .
That appears, to me, to not have any real roots.