Answer
$(-\infty,-\frac{1}{3}]\cup[5,\infty)$
Work Step by Step
Step 1. $3x^2\ge 14x+5 \Longrightarrow 3x^2-14x-5\ge0 \Longrightarrow (3x+1)(x-5)\ge0$
Step 2. Identify boundary points $x=-\frac{1}{3},5$ and form intervals $(-\infty,-\frac{1}{3}],[-\frac{1}{3},5],[5,\infty)$
Step 3. Choose test values $x=-1,0,6$, test the inequality to get results $True,\ False,\ True$.
Step 4. Thus we have the solution $(-\infty,-\frac{1}{3}]\cup[5,\infty)$
