In $\triangle ABC$, we have $AB = 14$, $BC = 16$, and $\angle A = 60^\circ$. Find the sum of all possible values of $AC$. In $\triangle ABC$, we have $AB = 14$, $BC = 16$, and $\angle A = 60^\circ$. Find the sum of all possible values of $AC$.
When I use the Law of Cosines, I get a quadratic like expected. However, when I use Vieta's Formulas to get the sum of the roots, I get the wrong answer. Please help me figure out what I did wrong.


Using the law of cosines we have,
$$\cos A=\frac{b^2+c^2-a^2}{2bc}$$
$$\cos\frac{\pi}{3}=\frac{b^2+196-256}{28b}$$
$$\frac12=\frac{b^2+196-256}{28b}$$
$$14b=b^2-60$$
$$b=7+\sqrt{109}$$
$$b\ne7-\sqrt{109}\mbox{ because the length cannot be in negative}$$
