jmc

Multiplying out each numerator and denominator, we get $$\frac{x^2-2x-3}{5(x^2-2x-8)}+\frac{x^2-2x-15}{9(x^2-2x-24)}-\frac{2(x^2-2x-35)}{13(x^2-2x-48)}=\frac{92}{585}.$$We can write this as $$\frac{(x^2-2x-8)+5}{5(x^2-2x-8)}+\frac{(x^2-2x-24)+9}{9(x^2-2x-24)}-\frac{2((x^2-2x-48)+13)}{13(x^2-2x-48)}=\frac{92}{585}.$$Hence, $$\frac{1}{5}+\frac{1}{x^2-2x-8}+\frac{1}{9}+\frac{1}{x^2-2x-24}-\frac{2}{13}-\frac{2}{x^2-2x-48}=\frac{92}{585}.$$This simplifies to $$\frac{1}{x^2-2x-8}+\frac{1}{x^2-2x-24}-\frac{2}{x^2-2x-48}=0.$$Let $y=x^2-2x-48.$ Then $$\frac{1}{y+40}+\frac{1}{y+24}-\frac{2}{y}=0.$$Multiplying everything by $y(y+24)(y+40),$ we get $$y(y+24)+y(y+40)-2(y+24)(y+40)=0.$$This simplifies to $64y+1920=0,$ so $y=-30.$ Then $x^2-2x-48=-30,$ or $x^2-2x-18=0.$ By the quadratic formula, $x=\boxed{1\pm \sqrt{19}}.$ (Since the denominators are nonzero for these values, we know that they are not extraneous.)