Linear meets Quadratic

If the parabola and the line above intersect at two distinct points, then we have the quadratic equation:

$x^2 + (a-19)x + (b+1) = 0 \Rightarrow x = \frac{(19-a) \pm \sqrt{(a-19)^2 - 4(1)(b+1)}}{2} = \frac{(19-a) \pm \sqrt{a^2 - 38a + 361 - 4b -4}}{2} = \frac{(19-a) \pm \sqrt{a^2 - 38a -4b + 357}}{2}$ .

If $x = 9-\sqrt{3}$ is one of the $x-$ coordinates of intersection, then we require:

$\frac{19-a}{2} = 9 \Rightarrow \boxed{a = 1}$ ;

$a^2 - 38a - 4b + 357 = 12 \Rightarrow \boxed{b = 77}$ ;

in order to attain $x = 9 \pm \sqrt{3}$ as the two distinct $x-$ coordinates of intersection. Hence, $\boxed{a+b=78}.$