12. Let $f(x)=4 x^{2}-4(a+1) x+3 a+3(a \in \mathbf{R})$, if the equation $f(x)=0$ has two distinct real roots both less than 2, is the inequality $(a+1) x^{2}-a x+a-1<0$ true for all real numbers $x$?
12. From the problem, we get $\left\{\begin{array}{l}\Delta=16(a+1)^{2}-16(3 a+3)>0 \\ \frac{a+1}{2}0,\end{array}\right.$ which yields $2<a<\frac{11}{5}$ or $a<-1$, and $\left\{\begin{array}{l}a+1<0 \\ a^{2}-4(a+1)(a-1)<0,\end{array}\right.$ which yields $a<-\frac{2 \sqrt{3}}{3}$. In summary, when $a<-\frac{2 \sqrt{3}}{3}$, $(a+1) x^{2}-a x+a-1<0$ holds for all real numbers $x$.