Multiplying both sides by $x+4$ and by $x+3$ gives
\[x(x+3) = -9(x+4).\]
Expanding both sides gives $x^2 + 3x = -9x -36$, and rearranging gives $x^2 + 12x + 36=0$. To factor this quadratic, we seek values $A$ and $B$ such that
\[x^2 + 12x + 36 = (x + A)(x + B) = x^2 + (A + B)x + AB,\]
so $A + B = 12$ and $AB = 36$. We find that $A = B = 6$ works, so
\[x^2 + 12x + 36 = (x + 6)(x + 6) = (x + 6)^2.\]
Therefore, the only solution is $x = \boxed{-6}$. We say that this is a "double root" of the quadratic.