jmc

We can pair the terms as follows: $$\left(\frac{1}{x}+\frac{1}{x+14}\right)+\left(\frac{1}{x+2}+\frac{1}{x+12}\right)-\left(\frac{1}{x+4}+\frac{1}{x+10}\right)-\left(\frac{1}{x+6}+\frac{1}{x+8}\right)=0.$$Then $$\frac{2x+14}{x^2+14x}+\frac{2x+14}{x^2+14x+24}-\frac{2x+14}{x^2+14x+40}-\frac{2x+14}{x^2+14x+48}=0.$$Dividing by 2, we get $$\frac{x+7}{x^2+14x}+\frac{x+7}{x^2+14x+24}-\frac{x+7}{x^2+14x+40}-\frac{x+7}{x^2+14x+48}=0.$$Let $y=x+7.$ Then $$\frac{y}{y^2-49}+\frac{y}{y^2-25}-\frac{y}{y^2-9}-\frac{y}{y^2-1}=0.$$We see that $y=0$ is a solution. Otherwise, $y\ne 0,$ so we can divide both sides by $y$: $$\frac{1}{y^2-49}+\frac{1}{y^2-25}-\frac{1}{y^2-9}-\frac{1}{y^2-1}=0.$$Now, let $z=y^2,$ so $$\frac{1}{z-49}+\frac{1}{z-25}-\frac{1}{z-9}-\frac{1}{z-1}=0.$$Then $$\frac{1}{z-49}-\frac{1}{z-9}=\frac{1}{z-1}-\frac{1}{z-25}.$$Combining the fractions on each side, we get $$\frac{40}{(z-49)(z-9)}=-\frac{24}{(z-1)(z-25)}.$$Hence, $40(z-1)(z-25)=-24(z-49)(z-9).$ This simplifies to $z^2-38z+181=0.$ By the quadratic formula, $$z=19\pm 6\sqrt{5}.$$Then $y=\pm \sqrt{19\pm 6\sqrt{5}},$ and $$x=-7\pm \sqrt{19\pm 6\sqrt{5}}.$$Thus, $a+b+c+d=7+19+6+5=\boxed{37}.$