jmc

Consider the quadratic formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. In order for the quadratic to have two non-real roots, the expression under the square root (the discriminant) must be negative. This gives us the inequality $$\begin{array}{rl}{b}^2-4ac& <0\\ \Rightarrow b^2-4(1)(9)& <0\\ \Rightarrow b^2-36& <0\\ \Rightarrow (b+6)(b-6)& <0.\end{array}$$ Thus, we find that $b\in \boxed{(-6,6)}$.