3. If the real roots $x_{1}, x_{2}$ of the equation $x^{2}+p x+q=0$ satisfy $x_{1}+x_{2}+x_{1} x_{2}>3$, then $p()$.
(A) is not less than 6
(B) is greater than 6 or less than -2
(C) is greater than -2 and less than 6
(D) is less than 6
3. (B)

From the problem, we get $p^{2} \geqslant 4 q,-p+q>3$, which means $4 q>4 p+12$. Solving the quadratic inequality $p^{2}>4 p+12$ yields $p>6$ or $p<-2$.