Answer
$\left( -\infty, -15 \right]\cup\left[ 1,\infty \right)$
Work Step by Step
Since for any $a\gt0$, $|x|\ge a$ implies $x\ge a\text{ or } x\le-a$, then the expression, $
\left| \dfrac{7+x}{2} \right|\ge4
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{7+x}{2}\ge4
\\\\
7+x\ge4(2)
\\\\
7+x\ge8
\\\\
x\ge8-7
\\\\
x\ge1
,\\\\\text{ OR }\\\\
\dfrac{7+x}{2}\le-4
\\\\
7+x\le-4(2)
\\\\
7+x\le-8
\\\\
x\le-8-7
\\\\
x\le-15
.\end{array}
Hence, the solution set is $
\left( -\infty, -15 \right]\cup\left[ 1,\infty \right)
.$
See the graph below.