Answer
$(-\infty,-2]U(1,\infty)$
Work Step by Step
Step 1. Graph $R(x)=\frac{2x+4}{x-1}$ as shown in the figure.
Step 2. The vertical asymptote ($x=1$) and a zero ($x=-2$) divide the x-axis into three intervals $(-\infty,-2],[-2,1),(1,\infty)$
Step 3. For $R(x)\ge0$, we can identify the solution intervals as $(-\infty,-2]U(1,\infty)$