If $(x+2)(x-3)=14$, find the sum of the possible values of $x$.
Expanding the left side of the given equation, we have $x^2-x-6=14 \Rightarrow x^2-x-20=0$. Since in a quadratic  with equation of the form $ax^2+bx+c=0$ the sum of the roots is $-b/a$, the sum of the roots of the given equation is $1/1=\boxed{1}$.