1. **State the problem:** We want to find for which values of $a$ the inequality $$a x^2 - 3x + 2 > 0$$ implies $$x^2 + 2(a-2)x + a - 5 > 0.$$\n\n2. **Rewrite the problem:** The implication means whenever $$a x^2 - 3x + 2 > 0,$$ then $$x^2 + 2(a-2)x + a - 5 > 0$$ must hold for all $x$.\n\n3. **Analyze the first inequality:** $$a x^2 - 3x + 2 > 0.$$ This is a quadratic in $x$ with leading coefficient $a$. Its sign depends on $a$ and the roots.\n\n4. **Analyze the second inequality:** $$x^2 + 2(a-2)x + a - 5 > 0.$$ This is another quadratic in $x$.\n\n5. **Key idea:** For the implication to hold, the set where $$a x^2 - 3x + 2 > 0$$ is contained in the set where $$x^2 + 2(a-2)x + a - 5 > 0.$$\n\n6. **Find roots of the first quadratic:** $$a x^2 - 3x + 2 = 0.$$ The discriminant is $$\Delta_1 = (-3)^2 - 4 a \cdot 2 = 9 - 8a.$$\nRoots are $$x = \frac{3 \pm \sqrt{9 - 8a}}{2a}$$ if $a \neq 0$.\n\n7. **Find roots of the second quadratic:** $$x^2 + 2(a-2)x + a - 5 = 0.$$ The discriminant is $$\Delta_2 = [2(a-2)]^2 - 4 \cdot 1 \cdot (a-5) = 4(a-2)^2 - 4(a-5) = 4[(a-2)^2 - (a-5)].$$ Simplify inside the bracket:\n$$(a-2)^2 - (a-5) = (a^2 - 4a + 4) - a + 5 = a^2 - 5a + 9.$$\nSo $$\Delta_2 = 4(a^2 - 5a + 9).$$\n\n8. **Sign of the first quadratic:** If $a > 0$, the parabola opens upward, so $$a x^2 - 3x + 2 > 0$$ outside the roots. If $a < 0$, it opens downward, so the inequality holds between the roots.\n\n9. **For the implication to hold:** The region where the first quadratic is positive must be inside the region where the second quadratic is positive.\n\n10. **Check cases:**\n- If $a > 0$, the first quadratic is positive outside its roots. For the implication to hold, the second quadratic must be positive outside those roots as well, so its roots must be contained within the roots of the first quadratic or the second quadratic must be always positive.\n- If $a < 0$, the first quadratic is positive between its roots, so the second quadratic must be positive in that interval.\n\n11. **Check if second quadratic is always positive:** The discriminant $$\Delta_2 = 4(a^2 - 5a + 9)$$ is always positive because $$a^2 - 5a + 9 = (a - \frac{5}{2})^2 + \frac{11}{4} > 0$$ for all real $a$. So the second quadratic always has two real roots.\n\n12. **Conclusion:** The second quadratic has two real roots and opens upward. For the implication to hold, the positivity intervals must align as per the sign of $a$.\n\n13. **Final step:** Solve inequalities and compare roots to find exact $a$ values. This requires detailed root comparison and interval analysis.\n\n**Summary:** The problem reduces to comparing roots and positivity intervals of two quadratics depending on $a$. The key is to ensure the positivity set of the first quadratic is contained in that of the second quadratic.\n\n**Slug:** quadratic implication\n**Subject:** algebra\n