Tomamos como el límite
$$\lim_{x \to 3^+}\left(\frac{2 x^{2} + \left(x - 15\right)}{7 x + \left(3 x^{2} - 6\right)}\right)$$
cambiamos
$$\lim_{x \to 3^+}\left(\frac{2 x^{2} + \left(x - 15\right)}{7 x + \left(3 x^{2} - 6\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{\left(x + 3\right) \left(2 x - 5\right)}{\left(x + 3\right) \left(3 x - 2\right)}\right)$$
=
$$\lim_{x \to 3^+}\left(\frac{2 x - 5}{3 x - 2}\right) = $$
$$\frac{-5 + 2 \cdot 3}{-2 + 3 \cdot 3} = $$
= 1/7
Entonces la respuesta definitiva es:
$$\lim_{x \to 3^+}\left(\frac{2 x^{2} + \left(x - 15\right)}{7 x + \left(3 x^{2} - 6\right)}\right) = \frac{1}{7}$$