smc

Let $y=m(x-20)+14$. Equating them: $x^2=mx-20m+14$ $x^2-mx+20m-14=0$ For there to be no solutions, the discriminant must be less than zero: $m^2-4(20m-14)<0$ $m^2-80m+56<0$. So $m<0$ for $r<m<s$ where $r$ and $s$ are the roots of $m^2-80m+56=0$ and their sum by Vieta's formulas is $\boxed{80}$.