#### Answer

$-6\lt a \lt0$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
25-2|a+3|\gt19
,$ isolate first the absolute value expression. Then use the definition of a less than (less than or equal to) absolute value inequality. Use the properties of inequality to isolate the variable. Finally, graph the solution set.
In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$
$\bf{\text{Solution Details:}}$
Using the properties of inequality to isolate the absolute value expression results to
\begin{array}{l}\require{cancel}
25-2|a+3|\gt19
\\\\
-2|a+3|\gt19-25
\\\\
-2|a+3|\gt-6
.\end{array}
Dividing both sides by a negative number (and consequently reversing the inequality symbol), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-2|a+3|\gt-6
\\\\
|a+3|\lt\dfrac{-6}{-2}
\\\\
|a+3|\lt3
.\end{array}
Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-3\lt a+3 \lt3
.\end{array}
Using the properties of inequality, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-3\lt a+3 \lt3
\\\\
-3-3\lt a+3-3 \lt3-3
\\\\
-6\lt a \lt0
.\end{array}
Hence, the solution set $
-6\lt a \lt0
.$