smc

We first start out by expanding the left side of the equation, $(x+2)(x+b)=x^2+bx+2x+2b=x^2+(2+b)x+2b=x^2+cx+6$. We know the constant terms have to be equal so we have $2b=6$, so $b=3$. Plugging $b=3$ back in yields $x^2+(2+3)x+6=x^2+cx+6$. Thus, $c=5⟹\boxed{5}$.