Problem 7. (15 points) Find all values of the parameter $a$ for which the equation $(a+2) x^{2}+(|a+3|-|a+11|) x+a=4$ has two distinct positive roots.
Solution: $(a+2) x^{2}+(|a+3|-|a+11|) x+a-4=0$.

When $a=-2$, the equation is linear and takes the form: $-8 x-6=0$, which does not satisfy the condition of the problem.

When $a \neq-2$, for the condition of the problem to be satisfied, it is necessary and sufficient that the following system is satisfied:

$$
\left\{\begin{array} { l } 
{ D > 0 } \\
{ - \frac { b } { a } > 0 } \\
{ \frac { c } { a } > 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(|a+3|-|a+11|)^{2}-4(a+2)(a-4)>0 \\
-\frac{(|a+3|-|a+11|)}{a+2}>0 \\
\frac{a-4}{a+2}>0
\end{array} \Leftrightarrow\right.\right.
$$

$\left\{\begin{array}{l}|a+3||a+11|0\end{array} \Leftrightarrow\left\{\begin{array}{l}a^{2}+14 a+33a^{2}-18 a-81 \\ \frac{(a+3)^{2}-(a+11)^{2}}{a+2}0\end{array} \Leftrightarrow\right.\right.$

$$
\left\{\begin{array}{l}
a^{2}-2 a-24-57 \\
\frac{2 a+14}{a+2}>0 \\
\frac{a-4}{a+2}>0
\end{array} \Leftrightarrow a \in(4 ; 6)\right.
$$

## Answer: $a \in(4 ; 6)$