jmc

Let $y=\sqrt{x-8},$ so $$\frac{6}{y-9}+\frac{1}{y-4}+\frac{7}{y+4}+\frac{12}{y+9}=0.$$Note that $$\frac{6}{y-9}+\frac{12}{y+9}=\frac{6(y+9)+12(y-9)}{y^2-81}=\frac{18y-54}{y^2-81}=\frac{18(y-3)}{y^2-81},$$and $$\frac{1}{y-4}+\frac{7}{y+4}=\frac{y+4+7(y-4)}{y^2-16}=\frac{8y-24}{y^2-16}=\frac{8(y-3)}{y^2-16},$$so $$\frac{18(y-3)}{y^2-81}+\frac{8(y-3)}{y^2-16}=0.$$If $y=3,$ then $x=3^2+8=17.$ Otherwise, we can divide both sides by $2(y-3),$ to get $$\frac{9}{y^2-81}+\frac{4}{y^2-16}=0.$$Multiplying both sides by $(y^2-16)(y^2-81),$ we get $$9(y^2-16)+4(y^2-81)=0.$$Then $13y^2=468,$ so $y^2=36.$ Since $y=\sqrt{x-8}$ must be nonnegative, $y=6.$ Then $x=6^2+8=44.$

Therefore, the solutions are $\boxed{17,44}.$