What is the value of $b+c$ if $x^2+bx+c>0$ only when $x\in (-\infty, -2)\cup(3,\infty)$ ?
By Vieta's formulas, b + c = (-2) + 3 + (-2)(3) = -5.
Figured it out myself, this is an interval problem not a quadratic problem, btw you were wrog the answer was -7.