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.

What you did wrong was forgetting that there is a negative solution!
When you use the Law of Cosines, you get
$$\begin{align}16^2 &= 14^2 + x^2 - 2(14)(x)cos(60^\circ) \\
        16^2 - 14^2 &= x^2 - 28x\cdot(\frac{1}{2}) \\
        16^2 - 14^2 &= x^2 - 14x \\
                 60 &= x^2 - 14x \\
                  0 &= x^2 - 14x - 60. \end{align}$$
Then we realize that one of the roots is negative leaving us with the answer
$$\boxed{7 + \sqrt{109}.}$$
