smc

Applying the difference of squares technique on this problem, we can see that $$(x-m{)}^2-(x-n{)}^2=((x-m)+(x-n))\cdot ((x-m)-(x-n)),$$ so $$(x-m{)}^2-(x-n{)}^2=((x-m)+(x-n))\cdot ((x-m)-(x-n))=(m-n{)}^2.$$ Simplifying gives us$$(2x-m-n)\cdot (n-m)=(m-n{)}^2.$$Negating $n-m$ creates:$$-(2x-m-n)\cdot (m-n)=(m-n{)}^2.$$Dividing by $m-n$, $$-(2x-m-n)=m-n$$ $$-2x+m+n=m-n$$ $$-2x+n=-n$$ $$-2x=-2n$$ $$x=n$$Lastly, since $n$ is a fixed negative number, $x$ must also be a fixed negative number, so $x<0$. Since this answer is not $A,B,C,$ or $D$, the solution must be $\boxed{\text{none of these}}$.