Answer
$\left( -\infty, -\dfrac{7}{6} \right) \cup \left( \dfrac{17}{6}, \infty\right)$
Work Step by Step
Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ or $x\lt-a,$ then the given inequality, $
|5-6x|\gt12
,$ is equivalent to
\begin{array}{l}\require{cancel}
5-6x\gt12 \text{ OR } 5-6x\lt-12
.\end{array}
Solving each inequality results to
\begin{array}{l}\require{cancel}
5-6x\gt12
\\\\
-6x\gt12-5
\\\\
-6x\gt7
\\\\
x\lt\dfrac{7}{-6}
\\\\
x\lt-\dfrac{7}{6}
\\\\\text{ OR }\\\\
5-6x\lt-12
\\\\
-6x\lt-12-5
\\\\
-6x\lt-17
\\\\
x\gt\dfrac{-17}{-6}
\\\\
x\gt\dfrac{17}{6}
.\end{array}
Hence, the solution is the interval $
\left( -\infty, -\dfrac{7}{6} \right) \cup \left( \dfrac{17}{6}, \infty\right)
.$
