1. **State the problem:** We are given the quadratic function $$f(x) = x^2 + 2px + p - 6$$ where $$p$$ is a constant. We need to find the values of $$m$$ and $$n$$ such that $$f(x)$$ is always positive when $$m < p < n$$. 2. **Recall the condition for a quadratic to be always positive:** A quadratic function $$ax^2 + bx + c$$ with $$a > 0$$ is always positive if its discriminant $$\Delta = b^2 - 4ac < 0$$. This means the quadratic has no real roots and opens upwards. 3. **Identify coefficients:** Here, $$a = 1$$, $$b = 2p$$, and $$c = p - 6$$. 4. **Calculate the discriminant:** $$\Delta = (2p)^2 - 4(1)(p - 6) = 4p^2 - 4p + 24$$ 5. **Set the discriminant less than zero:** $$4p^2 - 4p + 24 < 0$$ 6. **Simplify the inequality:** Divide both sides by 4: $$\cancel{4}p^2 - \cancel{4}p + \cancel{24} < 0 \Rightarrow p^2 - p + 6 < 0$$ 7. **Analyze the quadratic inequality:** The quadratic $$p^2 - p + 6$$ has discriminant: $$\Delta_p = (-1)^2 - 4(1)(6) = 1 - 24 = -23 < 0$$ Since the discriminant is negative, $$p^2 - p + 6$$ is always positive for all real $$p$$. 8. **Conclusion:** The inequality $$p^2 - p + 6 < 0$$ has no real solutions, so the discriminant $$\Delta$$ is never less than zero. This means the quadratic $$f(x)$$ is never always positive for any real $$p$$. However, the problem states $$f(x)$$ is always positive when $$m < p < n$$, so we must check if the quadratic can be positive for some interval. 9. **Check the leading coefficient:** Since $$a = 1 > 0$$, the parabola opens upwards. 10. **Check the vertex value:** The vertex $$x$$-coordinate is $$x_v = -\frac{b}{2a} = -\frac{2p}{2} = -p$$. The vertex value is: $$f(x_v) = (-p)^2 + 2p(-p) + p - 6 = p^2 - 2p^2 + p - 6 = -p^2 + p - 6$$ 11. **For $$f(x)$$ to be always positive, the vertex value must be positive:** $$-p^2 + p - 6 > 0$$ Multiply both sides by -1 (reverse inequality): $$p^2 - p + 6 < 0$$ But as shown before, $$p^2 - p + 6$$ is always positive, so this inequality has no solution. 12. **Therefore, the quadratic $$f(x)$$ is never always positive for any real $$p$$.** 13. **Check if the problem means always positive for some interval:** If the problem means $$f(x)$$ is positive for all $$x$$, then no such $$m,n$$ exist. If the problem means $$f(x)$$ is positive for some $$p$$ interval, then the problem might have a typo or different interpretation. 14. **Final answer:** No real values $$m,n$$ exist such that $$f(x)$$ is always positive for $$m < p < n$$. **Alternatively, if the problem meant always positive for all $$x$$, then the condition is no solution.**