1. **State the problem:** We want to find for which values of $a$ the inequality $$a x^2 + 2(a-2)x + a - 5 > 0$$ implies $$x^2 - 3x + 2 > 0.$$ 2. **Rewrite the inequalities:** The second inequality factors as $$x^2 - 3x + 2 = (x-1)(x-2) > 0.$$ This holds when $x < 1$ or $x > 2$. 3. **Analyze the first inequality:** The quadratic in $x$ is $$a x^2 + 2(a-2)x + a - 5 > 0.$$ For this to imply the second inequality, the set of $x$ values satisfying the first must be contained in the set where $x^2 - 3x + 2 > 0$. 4. **Find roots of the first quadratic:** The roots are given by $$x = \frac{-2(a-2) \pm \sqrt{4(a-2)^2 - 4a(a-5)}}{2a} = \frac{-2a + 4 \pm \sqrt{4(a-2)^2 - 4a^2 + 20a}}{2a}.$$ Simplify the discriminant: $$4(a-2)^2 - 4a^2 + 20a = 4(a^2 - 4a + 4) - 4a^2 + 20a = 4a^2 - 16a + 16 - 4a^2 + 20a = 4a + 16.$$ So roots are $$x = \frac{-2a + 4 \pm 2\sqrt{a + 4}}{2a} = \frac{-a + 2 \pm \sqrt{a + 4}}{a}.$$ 5. **Determine intervals where the first quadratic is positive:** Since the leading coefficient is $a$, if $a > 0$, the quadratic is positive outside the roots interval; if $a < 0$, it is positive inside the roots interval. 6. **Match intervals:** The first inequality's positive set must be contained in $(-\infty,1) \cup (2, \infty)$. - If $a > 0$, the positive set is $(-\infty, x_1) \cup (x_2, \infty)$ where $x_1 < x_2$ are roots. For this to be contained in $(-\infty,1) \cup (2, \infty)$, we need $x_1 \geq 1$ and $x_2 \leq 2$. - If $a < 0$, the positive set is $(x_1, x_2)$ and must be contained in $(-\infty,1) \cup (2, \infty)$, which is impossible since $(x_1, x_2)$ is a single interval. 7. **Check $a > 0$ case:** $$x_1 = \frac{-a + 2 - \sqrt{a + 4}}{a} \geq 1,$$ $$x_2 = \frac{-a + 2 + \sqrt{a + 4}}{a} \leq 2.$$ 8. **Solve inequalities:** For $x_1 \geq 1$: $$-a + 2 - \sqrt{a + 4} \geq a,$$ $$2 - \sqrt{a + 4} \geq 2a,$$ $$2 - 2a \geq \sqrt{a + 4}.$$ Square both sides (valid if RHS nonnegative and LHS nonnegative): $$(2 - 2a)^2 \geq a + 4,$$ $$4 - 8a + 4a^2 \geq a + 4,$$ $$4a^2 - 9a \geq 0,$$ $$a(4a - 9) \geq 0.$$ Since $a > 0$, this implies $a \geq \frac{9}{4} = 2.25$. For $x_2 \leq 2$: $$\frac{-a + 2 + \sqrt{a + 4}}{a} \leq 2,$$ $$-a + 2 + \sqrt{a + 4} \leq 2a,$$ $$\sqrt{a + 4} \leq 3a - 2.$$ For RHS to be nonnegative, $3a - 2 \geq 0 \Rightarrow a \geq \frac{2}{3}.$ Square both sides: $$a + 4 \leq (3a - 2)^2 = 9a^2 - 12a + 4,$$ $$0 \leq 9a^2 - 13a,$$ $$a(9a - 13) \geq 0.$$ Since $a > 0$, this implies $a \geq \frac{13}{9} \approx 1.444.$ 9. **Combine conditions:** From $x_1 \geq 1$, $a \geq 2.25$. From $x_2 \leq 2$, $a \geq 1.444$ and $a \geq \frac{2}{3}$ for RHS nonnegative. Overall, $a \geq 2.25$ satisfies both. 10. **Check $a=2.25$:** The conditions hold, so the solution is $$\boxed{a \geq \frac{9}{4}}.$$