#### Answer

$\left[ -\dfrac{11}{2},-\dfrac{1}{2} \right]$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
|-2x-6| \le 5
,$ use the definition of absolute value inequalities. Use the properties of inequalities to isolate the variable.
For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$
For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$
$\bf{\text{Solution Details:}}$
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}
-5 \le -2x-6 \le 5
.\end{array}
Using the properties of inequality, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-5+6 \le -2x-6+6 \le 5+6
\\\\
1 \le -2x \le 11
.\end{array}
Dividing by a negative number (and consequently reversing the sign), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1}{-2} \ge \dfrac{-2x}{-2} \ge \dfrac{11}{-2}
\\\\
-\dfrac{1}{2} \ge x \ge -\dfrac{11}{2}
\\\\
-\dfrac{11}{2} \le x \le -\dfrac{1}{2}
.\end{array}
In interval notation, the solution set is $
\left[ -\dfrac{11}{2},-\dfrac{1}{2} \right]
.$
The colored graph is the graph of the solution set.