jmc

Distribute on the left-hand side and add 3 to both sides to obtain $3x^2-7x+3=0$. Since it doesn't factor easily, we use the quadratic formula: $$\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{7\pm \sqrt{7^2-4\cdot 3\cdot 3}}{2\cdot 3}=\frac{7\pm \sqrt{13}}{6}.$$ Since $7$, $13$, and $6$ are relatively prime, $m=7$, $n=13$, and $p=6$, so $m+n+p=7+13+6=\boxed{26}$.