Answer
$\left( -\infty, -4 \right]\cup\left[ 1,\infty \right)$
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| 6+4x \right|\ge10
,$ is equivalent to
\begin{array}{l}\require{cancel}
6+4x\ge10
\\\\
4x\ge10-6
\\\\
4x\ge4
\\\\
x\ge\dfrac{4}{4}
\\\\
x\ge1
,\\\\\text{ OR }\\\\
6+4x\le-10
\\\\
4x\le-10-6
\\\\
4x\le-16
\\\\
x\le-\dfrac{16}{4}
\\\\
x\le-4
.\end{array}
Hence, the solution set is $
\left( -\infty, -4 \right]\cup\left[ 1,\infty \right)
.$