1. **Problem:** Solve the inequality $x^2 - 3x + 2 > 0$. 2. **Formula and rules:** For quadratic inequalities of the form $ax^2 + bx + c > 0$, first find the roots by solving $ax^2 + bx + c = 0$. The sign of the quadratic changes at the roots. Use the roots to test intervals. 3. **Find roots:** Solve $x^2 - 3x + 2 = 0$. $$x^2 - 3x + 2 = (x - 1)(x - 2) = 0$$ Roots are $x=1$ and $x=2$. 4. **Test intervals:** The roots divide the number line into three intervals: $(-\infty,1)$, $(1,2)$, and $(2,\infty)$. - For $x < 1$, pick $x=0$: $0^2 - 3\cdot0 + 2 = 2 > 0$ (True). - For $1 < x < 2$, pick $x=1.5$: $1.5^2 - 3\cdot1.5 + 2 = 2.25 - 4.5 + 2 = -0.25 < 0$ (False). - For $x > 2$, pick $x=3$: $9 - 9 + 2 = 2 > 0$ (True). 5. **Solution:** The inequality holds for $x \in (-\infty,1) \cup (2,\infty)$. **Final answer:** $$x \in (-\infty,1) \cup (2,\infty)$$