If the solution set of the quadratic inequality $ax^2+bx+c>0$ is $(2,3)$, find the solution set for $cx^2+bx+a<0$
If the set where $ax^2+bx+c > 0$ is $(2,3)$, the only possibility is that $$ax^2+bx+c = -(x-2)(x-3)$$
So $$ax^2+bx+c = -x^2 +5x-6$$
So $a=-1$, $b=5$ and $c=-6$.
Therefore you want to solve $-6x^2 +5x-1 < 0$. You can easily see that the roots of the new polynomials are $1/2$ and $1/3$, and therefore the set where $-6x^2 +5x-1 < 0$ is $(-\infty, 1/3) \cup (1/2, +\infty)$.