已知a=5

由已知得（a+1）²=5，所以a²+2a=4
则原式=2a³+4a²+2a+3a²-4a-12
=2a（a²+2a+1）+3a²-4a-12
=2a（a+1）²+3a²-4a-12
=2a×5+3a²-4a-12
=3a²+6a-12
=3（a²+2a）-12
=3×4-12
=0
故答案0