Optical isomerism in Co-ordination Compounds
Any 3rd-year inorganic chemistry textbook worth its weight in salt will provide some information about isomers in coordination compounds.  An optically active compound contains no rotation-reflection symmetry element.  Typically, this means that if a molecule has an inversion center or a mirror plane then it is not chiral.  

It takes a systematic brute force and ignorance approach to determining the number of isomers a coordination compound has, and then a bit more visual thinking to determine if any of the isomers are enantiomers.  The following figure is taken from Miessler, Fischer, Tarr 5th edition and is useful for identifying isomers of 6-coordinate complexes.



Note that a 6-coordinate complex with 6 unique ligands in an octahedral field would be enantiomers. In fact, there are 15 different isomers, each of which is chiral resulting in a total of 30 isomers!

The method of determining isomers this way is to formulate a structure of the type M where  indicates two ligands that are trans to each other.  Then, you start varying a, b, c, ... until you've come up with as many unique combinations as possible.  For your case, you have 2 instances of a bidentate ligand.  Bidentate ligands are typically designated with capital letters in this notation, so the formula for your molecule would be $\ce{M(AA)_2b_2}$.  Further, since bidentate ligands are too small to coordinate in a trans conformation, you will decrease the number of possible isomers.  One isomer, for example, would be M, where I use A and A' to designate the different $\ce{(en)}$ ligands.

The answer to your question, then, is that there are two isomers, and they are enantiomers, since you can't rearrange the A, A' and b in the formula above to come up with another unique configuration.