I have a question which asks...

What is the Oxidation number of Nitrogen in Cu(NO3)2
I know the answer is 5+ but how do you get that??

Cu= 2+
O= 2- (Theres 3 of them so -2x3=-6...then, theres a two out of the brackets so 6- x 2 =-12)
+2 - (-12)= -10? And then what?
I think my method is wrong. :/ It's really frustrating and i've been at the same question for over an hour now.

I do it this way. First, do everything except the element of interest.

Cu +2
6 O = -12
The only thing left is N. So for the N in the formula, we have -12+2 = -10; therefore, N must be +10 to make everything zero. There are 2 N atoms @ +10 total which makes each N = +5.