2014 AMC 12B Problem 17

The line through $Q$ is $y = m(x-20) + 14.$ Substituting into $y = x^2$ gives $$x^2 - mx + (20m - 14) = 0.$$ There is no intersection exactly when this has no real root, i.e. when the discriminant $m^2 - 4(20m-14)$ $= m^2 - 80m + 56$ is negative. That happens between the two roots $r$ and $s$ of $m^2 - 80m + 56 = 0.$ By Vieta’s formulas, $r + s = 80.$ Thus, the correct answer is E.