Question 738977
 Find all possible numbers where twice the reciprocal of the square of a number less three times the reciprocal of a number is -1. 
:
{{{2/x^2}}} - {{{3/x}}} = -1
multiply eq by x^2, canceling the denominators, resulting in:
2 - 3x = -x^2
Combine as a quadratic equation on the left
x^2 - 3x + 2 = 0
Factors to
(x-1)(x-2) = 0
Two solutions
x = 1
and 
x = 2