Answer
$(-\infty,-2]\cup\left[ \frac{3}{2},\infty\right)$.
Refer to the graph below.
Work Step by Step
Step $1$. Remove the absolute value sign using the rule $|a|\ge b\longrightarrow a\ge b \text{ or } a\le b$ to obtain:
$$4x+1\ge7\quad \text{or} \quad 4x+1\le -7$$
Step $2$. Solve each inequality.
$$\begin{align*}
4x+1&\ge7\\
4x&\ge6\\
x&\ge \frac{3}{2}
\end{align*}$$
$$\begin{align*}
4x+1&\le-7\\
4x&\le-8\\
x&\le -2
\end{align*}$$
Thus, the solution is $(-\infty, -2] \cup \left[\frac{3}{2}, \infty\right)$.
Step $3$. Graph the solution by plotting a bracket at $-2$ and shading the region to its left, then plotting a bracket at $\frac{3}{2}$ then shading the region to its right. Refer to the image below.