Answer: 13

Small Hint:

Set $y=x^2-10x$ so the three denominators differ only by constants

Big Hint:

Combine the first two fractions before clearing denominators

Solution:

Set $y=x^2-10x.$ Combining the first two fractions and clearing the nonzero denominators gives an equality between $(y-37)(y-69)$ and $(y-29)(y-45).$ Expanding and canceling $y^2$ yields $y=39.$ Thus $x^2-10x-39=0,$ so $(x-13)(x+3)=0.$ The positive solution is $13.$