3a³-6a²-14a+20
=3a³-6a²-24a+10a+20
=3a(a²-2a-8)+10(a+2)
=3a(a-4)(a+2)+10(a+2)
=(a+2)(3a²-12a+10)=0
解得a=-2或者a=(4±√6)/3