jmc

If the given equation is true, then multiplying by $(x+C)(x+7)$ gives the equation $$(x+B)(Ax+28)=2(x+C)(x+7),$$which must also be true. (Note however, that the converse does not hold: that is, by multiplying by $(x+C)(x+7),$ we may have introduced extraneous roots.) Therefore, the above equation must also have infinitely many roots for $x.$ That is, the polynomials $(x+B)(Ax+28)$ and $2(x+C)(x+7)$ must agree for infinitely many values of $x.$ This means that they must be identical polynomials. (In general, if $p(x)=q(x)$ for infinitely many $x,$ then $p(x)-q(x)=0$ has infinitely many roots, which is only possible if $p(x)-q(x)$ is identically the zero polynomial.)

This means that $$(x+B)(Ax+28)=2(x+C)(x+7)$$for all $x.$ Expanding both sides, we get $$A{x}^2+(AB+28)x+28B=2x^2+(2C+14)x+14C.$$Corresponding coefficients of both sides must be equal, so we have $$\begin{array}{rl}A& =2,\\ AB+28& =2C+14,\\ 28B& =14C.\end{array}$$From the first and third equations, $A=2$ and $C=2B.$ Then substituting into the second equation gives $$2B+28=4B+14,$$so $B=7,$ and then $C=14.$ This means that our original equation was $$\frac{(x+7)(2x+28)}{(x+14)(x+7)}=2.$$This equation holds whenever the denominator is nonzero. The denominator is equal to zero when $x=-7$ and $x=-14,$ so the sum of the values of $x$ which are not roots of the original equation is $(-7)+(-14)=\boxed{-21}.$