How would u solve
D\d+2 - 2\2-d = d+6\ d^2-4
I got to
D(d-2)(2-d) -2 (d+2)(d-2)=2d-d^2+12-6d
How would you solve? Im confused
Did you not notice that the denominator of the right side is (d-2)(d+2) ?
I will also assume that D is really d, or else you have 2 unknowns.
so let's rewrite ...
d/(d+2) - 2/(2-d) = (d+6)/[(d+2)d-2)]
d/(d+2) + 2/(d-2) = (d+6)/[(d+2)d-2)]
now multiply each term by (d+2)(d-2)
d(d-2) + 2(d+2) = d+6
d^2 - 2d + 2d + 4 = d + 6
d^2 - d - 2 = 0
(d-2)(d+1) = 0
d = 2 or d = -1