Answer
$(-\infty,\infty)$ or $\{x|x=all\ real\ numbers\}$.
Work Step by Step
Step 1. $2(2x^2-3x)\ge-9 \longrightarrow 4x^2-6x+9\ge0 \longrightarrow b^2-4ac=(-6)^2-4(4)(9)=-108\lt0$.
Step 2. Thus the quadratic opens up and has no x-intercept, meaning the curve is above the x-axis and the solution interval is $(-\infty,\infty)$ or $\{x|x=all\ real\ numbers\}$.
