Following Aleksey Pichugin's argument, the question is to find the condition in which $px^2+(p-3)x+(2-p)=0$ ($\ast$) has at least one positive root.
First, its discriminant must be non-negative: $(p-3)^2-4p(2-p)=(p-1)(5p-9)\ge 0$, which means $p\le 1$ or $p\ge 9/5$. (1)
Provided (1) is satisfied, the negation of 'the equation ($\ast$) has at least one positive root' is 'the equation ($\ast$) has two nonpositive (=negative or zero) roots'. Let's consider the latter condition, which seems a bit easier.
If $p=0$, ($\ast$) reduces to $-3x+2=0$ and it has a positive root; we assume $p\ne 0$ below. Let the two roots of the equation be $\alpha$ and $\beta$ (provided (1) is satisfied). Then '$\alpha\le 0$ and $\beta\le 0$' is equivalent to '$\alpha\beta\ge 0$ and $\alpha+\beta\le 0$'. Here $\alpha\beta=\frac{2-p}{p}$, while $\alpha+\beta=-\frac{p-3}{p}$, so this means '$p(2-p)\ge 0$ and $p(3-p)\le 0$'. There are no $p\ne 0$ satisfying both inequalities.
Therefore the condition we want to have is (1), which is '$p\le 1$ or $p\ge 9/5$'.