jmc

Distribute on the left-hand side and subtract 3 from both sides to obtain $2x^2-7x-3=0$. Inspection reveals that $2x^2-7x-3$ does not easily factor, so we substitute the coefficients $2$, $-7$, and $-3$ into the quadratic formula: $$\frac{-(-7)\pm \sqrt{(-7{)}^2-(4)(2)(-3)}}{2(2)}=\frac{7\pm \sqrt{49+24}}{4}=\frac{7\pm \sqrt{73}}{4}.$$Therefore $m=7$, $n=73$, and $p=4$, so $m+n+p=7+73+4=\boxed{84}$.