Answer
The inequality is valid for all values i.e. $(-\infty,\infty)$
Work Step by Step
First, we are going to set the right side to zero and use the quadratic equation to find the x-intercepts:
$2(2x^2-3x)=-9$
$4x^2-6x+9=-9+9$
$4x^2-6x+9=0$
a=4, b=-6, c=9
$x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(4)(9)}}{2(4)}$
$x=\dfrac{6\pm\sqrt{36-144}}{8}$
We can see that there will be a negative value in the radical, which means the equation doesn't have real solutions and that there aren't x-intercepts.
Since the equation never crosses the x-axis and is upward facing (all f(x) values are positive), the inequality $2(2x^2-3x)>-9$ is valid for all values i.e. $(-\infty,\infty)$
