jmc

First we bring $3x$ to the left side to get $$x^2-7x-14=16.$$Moving the 14 to the right side gives $$x^2-7x=30.$$We notice that the left side is almost the square ${\left(x-\frac{7}{2}\right)}^2=x^2-7x+\frac{49}{4}$. Adding $\frac{49}{4}$ to both sides lets us complete the square on the left-hand side, $$x^2-7x+\frac{49}{4}=30+\frac{49}{4}=\frac{169}{4},$$so $${\left(x-\frac{7}{2}\right)}^2={\left(\frac{13}{2}\right)}^2.$$Therefore $x=\frac{7}{2}\pm \frac{13}{2}$. The positive difference between these solutions is $$\frac{7+13}{2}-\frac{7-13}{2}=\frac{26}{2}=\boxed{13}.$$