#### Answer

$(-\infty, -2] \cup [0.5, +\infty)$
Refer to the attached image below for the graph.

#### Work Step by Step

Subtract $3x^2+2$ to both sides to have:
$\\5x^2+3x-(3x^2+2) \ge 3x^2+2 - (3x^2+2)
\\2x^2+3x-2 \ge 0$
Factor the trinomial to have:
$\\(2x-1)(x+2)\le0$
Now that the nonzero side of the inequality is already factored, proceed to finding the intervals for the given inequality.
The factors of the nonzero side are $2x-1$ and $(x+2)$.
These factors are zero when $x=0.5 \text{ and } -2$, respectively.
These numbers divide the number line into the following intervals:
$(−\infty, -2),(-2, 0.5), (0.5, +\infty)$
Make a diagram by using test points to determine the sign of each factor in each interval. (refer to the attached image below).
It can be seen from the diagram that the given inequality is satisfied on the intervals $(-\infty, -2) \text{ and } (0.5, +\infty)$.
The inequality involves $\ge$ so the endpoints satisfy the inequality.
Thus, the solution is $ (-\infty, -2] \cup [0.5, +\infty)$.