Answer
The solution set is $(-\infty,-1]\cup[\frac{1}{2},\infty)$
Work Step by Step
$2x^2+x\ge1\hspace{0.7cm}{\color{blue}{\text{Given equation}}}$
$\Rightarrow 2x^2+x-1\ge0$
$\Rightarrow (2x-1)(x+1)\ge0\hspace{0.7cm}{\color{blue}{\text{Factor}}}$
The factors of the left-hand side are $2x-1$ and $x+1$.
These factors are zero when $x=\frac{1}{2}$ and $x=-1$.
These numbers divide the real line into the intervals
$(-\infty,-1),(-1,\frac{1}{2}),(\frac{1}{2},\infty)$
From the diagram and hence the inequality involves $\ge$, the endpoints of the intervals satisfy the inequality.
The solution set is $(-\infty,-1]\cup[\frac{1}{2},\infty)$
