Tomamos como el límite
$$\lim_{x \to -4^+}\left(\frac{7 x + \left(x^{2} + 9\right)}{2 x^{2} + \left(- 5 x - 3\right)}\right)$$
cambiamos
$$\lim_{x \to -4^+}\left(\frac{7 x + \left(x^{2} + 9\right)}{2 x^{2} + \left(- 5 x - 3\right)}\right)$$
=
$$\lim_{x \to -4^+}\left(\frac{x^{2} + 7 x + 9}{\left(x - 3\right) \left(2 x + 1\right)}\right)$$
=
$$\lim_{x \to -4^+}\left(\frac{x^{2} + 7 x + 9}{\left(x - 3\right) \left(2 x + 1\right)}\right) = $$
$$\frac{\left(-4\right) 7 + 9 + \left(-4\right)^{2}}{\left(-4 - 3\right) \left(\left(-4\right) 2 + 1\right)} = $$
= -3/49
Entonces la respuesta definitiva es:
$$\lim_{x \to -4^+}\left(\frac{7 x + \left(x^{2} + 9\right)}{2 x^{2} + \left(- 5 x - 3\right)}\right) = - \frac{3}{49}$$