#### Answer

$\left( -\infty, -4 \right) \cup \left( 7, \infty\right)$

#### Work Step by Step

Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ or $x\lt-a,$ then the given inequality, $
|2x-3|\gt11
,$ is equivalent to
\begin{array}{l}\require{cancel}
2x-3\gt11 \text{ OR } 2x-3\lt-11
.\end{array}
Solving each inequality results to
\begin{array}{l}\require{cancel}
2x-3\gt11
\\\\
2x\gt11+3
\\\\
2x\gt14
\\\\
x\gt\dfrac{14}{2}
\\\\
x\gt7
\\\\\text{ OR }\\\\
2x-3\lt-11
\\\\
2x\lt-11+3
\\\\
2x\lt-8
\\\\
x\lt-\dfrac{8}{2}
\\\\
x\lt-4
.\end{array}
Hence, the solution is the interval $
\left( -\infty, -4 \right) \cup \left( 7, \infty\right)
.$