The set of all possible values of $AC$ is an open interval $(m,n)$ 
Side $\overline{AB}$ of $\triangle ABC$ has length $10$. The bisector of angle $A$ meets $\overline{BC}$ at $D$, and $CD = 3$. The set of all possible values of $AC$ is an open interval $(m,n)$. What is $m+n$?
A) 16 B) 17 C) 18 D) 19 E) 20

Could someone help me understand what is wrong with my approach (none of the answers matches mine!):
Letting BD equal $x$ (which must be positive) and using the angle bisector theorem, we get that AC is $\frac{30}{x}$. Using triangular inequality, we get that $$7 < x+\frac{30}{x}$$ $$x<7+\frac{30}{x}$$ and $$\frac{30}{x} < 13+x.$$
We can multiply $x$ to both sides of all 3 inequalities because $x>0$ and will not change the direction of the inequality. Doing so and moving all terms to the LHS, we get $x^2-77x+30>0$ (1), $x^2-7x-30<0$ (2) and $x^2+13x-30>0$ (3). (1) is always true because we can re-write it in the form of $(x-a)^2+b>0$ where $a$ and $b$ are both positive. (2) can be rewritten as $(x+3)(x-10)<0$, and thus $-3<x<10$. But $x>0$, so $0<x<10$.
Finally, (3) can also be written as $(x+15)(x-2)>0$ where only $x>2$ works because $x<-15$ is less than 0. Then combining all the scenarios, $x$ can range from $2$ to $10$, but $12$ is not an option... I've double/triple checked my work, but there must be something I'm missing.

I believe you have the correct range for $x$ of $2 \lt x \lt 10$, but that is for $\overline{BD}$. However, the question asks for $\overline{AC}$. Since $\overline{AC} = \frac{30}{x}$, this means it's range is $3 \lt \overline{AC} \lt 15$, so $m = 3$ and $n = 15$, giving the correct answer of $m + n = 3 + 15 = 18$, i.e., choice $C$.
