is he or is he just another good and gone qb.
can he really take them to the playoffs and win there. or even take them to the super bowl.
2006-11-27 13:15:22 · 11 answers · asked by Anonymous in Sports ➔ Football (American)
An undrafted free agent he could be. If he can establish himself Bledsoe will want his ticket punched. Next stop, post season.
2006-11-27 15:22:27 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I've had a good feeling about Romo and i think he may take the cowboys into far playoffs...maybe all the way, but i think he will stop short in 3-4 years. you never know until it happens.
2006-11-27 21:23:21 · answer #2 · answered by Anonymous · 0⤊ 0⤋
No, Romo is a stop-gap QB. He'll be there for 2-3 yrs but I wouldn't call him a franchise QB ever. Think Steve McNair, not Steve Young.
2006-11-27 21:18:34 · answer #3 · answered by jesser312 2 · 0⤊ 0⤋
He is for the time being. The Cowboys will win the division, the eagles suck and the little giants are beating themselves so GO COWBOYS!!!!!!!!!!!
2006-11-27 21:40:23 · answer #4 · answered by Scottie Boy 6 · 0⤊ 0⤋
looks like he is the answer and could easily take them to the playoffs and maybe even go to the superbowl.
2006-11-27 21:51:57 · answer #5 · answered by mister ss 7 · 0⤊ 0⤋
it looks like he is, as long as he keeps playing like that he will be there. They are already comparing him to Aikman, which I think it is a little to soon for that, let him at leastmake the playoffs before you compare him to Aikman.
2006-11-27 22:02:44 · answer #6 · answered by rcaines1 3 · 0⤊ 0⤋
he has been most of the reason why the cowboys are playing better, but the secondary has also stepped up and they are making a lot fewer penalites.
2006-11-27 21:33:22 · answer #7 · answered by HITMAN 3 · 0⤊ 0⤋
With out a doubt, yes.
2006-11-27 21:26:55 · answer #8 · answered by Jazzy 5 · 0⤊ 0⤋
yes i think hes better then bletso
2006-11-27 21:24:18 · answer #9 · answered by Anonymous · 0⤊ 0⤋
yeah but they will never beat the EAGLES!
2006-11-27 21:22:57 · answer #10 · answered by meg0xx 2 · 0⤊ 1⤋