Example 5. One root of the equation $3 x^{2}-5 x+a=0$ is greater than -2 and less than $0_{3}$, the other is much greater than 1 and less than 3. Find the range of real values for a.
Solve: The equation can be rewritten as
$$
\mathrm{x}^{2}-\frac{5}{3} \mathrm{x}+\frac{\mathrm{a}}{3}=0 \text {. }
$$

Here $\mathrm{p}=-\frac{5}{3}, \mathrm{q}=\frac{\mathrm{a}}{3}$. According to Theorem 7, the necessary and sufficient condition for one root of this equation to be greater than -2 and less than 0, and the other root to be greater than 1 and less than 3 is
$$
\left\{\begin{array}{l}
(-2)^{2}+\mathrm{p} \cdot(-2)+\mathrm{q}=\frac{22}{3}+\frac{\mathrm{a}}{3}>0 . \\
0^{2}+\mathrm{p} \cdot 0+\mathrm{q}=-\frac{\mathrm{a}}{3}<0,
\end{array}\right.
$$

Solving this system of inequalities yields $-12<\mathrm{a}<0$.