What is the value of b + cif x^2 + bx + c > 0 if only when $x \in (-\infty,2) \cup (3,\infty)$?
The function will be negative on the interval (2, 3)
This implies that we have
(x - 2) ( x - 3) > 0
x^2 - 5x + 6 > 0
b = -5 c = 6
b + c = 1
