Scotch Gambit, 4...Bb4+ Variation
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1.e4 e5 2.Nf3 Nc6 3.d4 exd4 4.Bc4 Bb4+ 5.c3 dxc3 I prefer bxc3. After Ba5, O-O. A typical line might be d6 Qb3 Qe7 e5 dxe5 Ba3 Qf6 Nbd2 (heading for e5) etc.
Scotch gambit is only a temporary gambit in most lines. The bishop check line is the one line that ensures that white will be a pawn down with no immediately obvious way to recover the material. I feel like black already has an advantage within the first 10 moves with not enough compensation for white.
@mikeploch, your diagram is a good one. But i would like to add some more details into it. For example, in the 7...Nge7 sideline, you claims that 'Black's in trouble after f4'. Well, instead of 9...h6, there's 9...Bc5, pinning that f-pawn. White may regain his pawn back after 10.Nxf7(10.Kh1 unpinning the f-pawn doesn't work at all after 10...d5! 11.exd5 O-O/Bg4) Nxf7 11.Bxf7+ Kxf7 12.Qh5+ g6 13.Qxc5 d6, winning some tempos on White's queen, White's only developed piece atm.
Black has also other alternatives on move 7, e.g, 7...Qe7 with ...Nf6 idea, or even the immediate 7...Nf6?! 8.e5!? Bxe5 9.Nxe5 Nxe5 10.Re1 d6 11.f4 O-O 12.fxe5 dxe5 13.Qxd8 Rxd8 14.Bg5(14.Rxe5? Rd1+ wins a piece back) h6 15.Bxf6 gxf6, with 3p for a B. Not bad at all, huh? 
The issue with the 9 ...Bc5 line is that Black's King makes his position pretty awkward. Nxf7 Nxf7 Bxf7+ Kxf7 Qh5+ Ng6 Qd5+ Kf8 Qf5+ forces Black to choose between a useful Rook (with ...Ke8) or a safer King (with ...Kg8).
On 7 ...Nf6 e5 Bxe5 Nxe5 Nxe5 Re1 d6 f4 0-0 fxe5 dxe5 Qxd8 Rxd8 Fritz recommends Nd2 over Bg5. The idea is that Black's e5-pawn is not a strong one. Black does get three pawns for the piece, but I don't think he can hold them.
In most of the games I've saw with the Scotch Gambit(1.e4 e5 2.Nf3 Nc6 3.d4 exd4 4.Bc4), Black tends to play the more common 4...Bc5 or 4...Nf6.
But there's another move, i.e, 4...Bb4+, on which simply develops a piece with a gain of tempo by attacking White's king, similar to the Danish Gambit Accepted(1.e4 e5 2.d4 exd4 3.c3 dxc3 4.Bc4 cxb2 5.Bxb2 Bb4+), covered by IM pfren in one of his articles.
For instance, 5.Bd2 is met by 5...Bxd2+ 6.Qxd2 Nf6 7.e5 Ng4 8.Bxf7+(8.Qf4 O-O, since 9.Qxg4 d5 wins the piece back) Kxf7 9.Qf4+ Nf6 10.exf6 Re8+ followed by Qxf6.
And 5.c3 dxc3 6.O-O cxb2 7.Bxb2 Nf6 8.Ng5 O-O 9.e5 d5 10.exf6 dxc4 11.Qh5(11.fxg7 Qxg5 12.gxf8=Q+ Kxf8 and Black has at least equality material(B+2P vs R) and positional) h6 12.Ne4 Re8 13.Rd1 Bd7(13...Qxd1+ 14.Qxd1 Rxe4 is acceptable too) 14.fxg7 Rxe4 15.Qxh6 f5(15...Rh4?? 16.Qh8+ Rxh8 17.gxh8=R(Q)#) and white will have perpetual check at best.
Comments/improvements are welcome.