Are you smarter than me? 38

Let $(x-3)(x-4)=\dfrac {34}{1089}\quad \Rightarrow x^2 - 7x + 12 - \dfrac {34}{1089} = 0$

Therefore, the larger root of the equation is:

$\dfrac {7 + \sqrt{7^2 - 4\left( 12 - \dfrac {34}{33^2}\right)} } {2} = \dfrac {7}{2} + \dfrac {\sqrt{49(33^2) - 48(33^2) + 4(33+1)} }{66}$

$= \dfrac {7}{2} + \dfrac {\sqrt{33^2 + 4(33) +4}}{66} = \dfrac {7}{2} + \dfrac {\sqrt{(33+2)^2}}{66}$

$= \dfrac {7}{2} + \dfrac {35}{66} = \dfrac {231 + 35} {66} = \dfrac {266} {66} = \dfrac {133} {33} = \dfrac {m}{n}$

$\Rightarrow m = 133$ and $n = 33\quad \Rightarrow m+n = \boxed{166}$

assume (x-3)=m therefore m(m-1)=34/33^2

solving quadratic by splitting the middle term

m^2-m-34/33^2=0 =(m-34/33)(m+1/33)

further solving we get x=133/33=166

$(x-3)(x-4)=\dfrac{34}{1089}~~~~~~~~\implies~x^2-7x-12-\dfrac{34}{1089}=0\\ x = \dfrac{7 +\sqrt{49 - 4*(12- \dfrac{34}{1089} )} }{2} \\ x= \dfrac{7 +\sqrt{1 +4* \dfrac{34}{1089} } }{2}=\dfrac{7 +\sqrt{ \dfrac{1225}{1089} } }{2}\\x=\dfrac{ 7 + \dfrac{35}{33} }{2}\\x=\dfrac{ 8 + \dfrac{2}{33} }{2}~~~~ ~~~~~~x=4+\dfrac{1}{33}=\dfrac{133}{33} =\dfrac{m}{n} \\ \therefore~m+n= \boxed {166 }$