Integers $< 100$

Taking the RHS inequality, we get $x^2-14x+11 < x^2-2x+3 \Rightarrow x > \frac{2}{3}.$ For the LHS inequality, we have $-x^2 + 2x-3 < x^2 -14x +11 \Rightarrow 0 < 2x^2 -16x+14 \Rightarrow 0 < x^2 -8x+7 \Rightarrow 0 < (x-1)(x-7) \Rightarrow x \in (-\infty,1) \cup (7,\infty)$ . This gives us the entire solution set $x \in (\frac{2}{3},1) \cup (7,\infty)$ . If we want $x \in \mathbb{N}$ over the interval $(7,100)$ , then there are $\boxed{92}$ admissible values.