#### Answer

$x\le-15
\text{ OR }
x\ge9$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
|x+3|\ge12
,$ use the definition of 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 $\ge.$
$\bf{\text{Solution Details:}}$
Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
x+3\ge12
\\\\\text{OR}\\\\
x+3\le-12
.\end{array}
Solving each inequality results to
\begin{array}{l}\require{cancel}
x+3\ge12
\\\\
x\ge12-3
\\\\
x\ge9
\\\\\text{OR}\\\\
x+3\le-12
\\\\
x\le-12-3
\\\\
x\le-15
.\end{array}
Hence, the solution set is $
x\le-15
\text{ OR }
x\ge9
.$