1. **Problem:** Simplify $$\frac{x^2-25}{x^2+5x}\div\frac{xy+6x-5y-30}{5x-15}$$ 2. **Use the division rule for fractions:** $$\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\cdot\frac{d}{c}$$ So we rewrite the problem as $$\frac{x^2-25}{x^2+5x}\cdot\frac{5x-15}{xy+6x-5y-30}$$ 3. **Factor each expression:** $$x^2-25=(x-5)(x+5)$$ $$x^2+5x=x(x+5)$$ $$5x-15=5(x-3)$$ $$xy+6x-5y-30=x(y+6)-5(y+6)=(x-5)(y+6)$$ 4. **Substitute the factored forms:** $$\frac{(x-5)(x+5)}{x(x+5)}\cdot\frac{5(x-3)}{(x-5)(y+6)}$$ 5. **Cancel common factors carefully:** $$\frac{(x-5)\cancel{(x+5)}}{x\cancel{(x+5)}}\cdot\frac{5(x-3)}{\cancel{(x-5)}(y+6)}$$ 6. **Multiply what remains:** $$\frac{5(x-3)}{x(y+6)}$$ 7. **Final answer:** $$\boxed{\frac{5(x-3)}{x(y+6)}}$$