$$x^2+6x+9=\left|x+3\right|$$ The absolute value gives us two options: $$x^2+6x+9=x+3$$ or $$x^2+6x+9=-\left(x+3\right)$$ $$x^2+6x+9=-x-3$$ Solving each individually:
$$x^2+6x+9=x+3$$ $$x^2+5x+6=0$$ $$\left(x+2\right)\left(x+3\right)=0$$ $$x=-2\ and\ x=-3$$
and $$x^2+6x+9=-x-3$$ $$x^2+7x+12=0$$ $$\left(x+3\right)\left(x+4\right)=0$$ $$x=-3\ and\ x=-4$$
Notice that x=-3 is a solution for both equations. We are looking for total *unique* solutions, so we'll only include it once.
So our unique solutions are -2, -3, and -4. To find the sum, we'll add them together: $$-2\ +\left(-3\right)+\left(-4\right)=-2-3-4=-9$$ This gives B as the correct answer.