1. **State the problem:** Solve the inequality $$-2x^2 + 9x - 17 > -8x - 9$$ algebraically. 2. **Bring all terms to one side:** Add $$8x + 9$$ to both sides to get $$-2x^2 + 9x - 17 + 8x + 9 > 0$$ 3. **Combine like terms:** $$-2x^2 + (9x + 8x) + (-17 + 9) > 0$$ $$-2x^2 + 17x - 8 > 0$$ 4. **Rewrite the inequality:** $$-2x^2 + 17x - 8 > 0$$ 5. **Multiply both sides by $$-1$$ to make the quadratic coefficient positive, remembering to reverse the inequality sign:** $$\cancel{-1} \times (-2x^2 + 17x - 8) < \cancel{-1} \times 0$$ $$2x^2 - 17x + 8 < 0$$ 6. **Factor the quadratic:** Find two numbers that multiply to $$2 \times 8 = 16$$ and add to $$-17$$. These are $$-16$$ and $$-1$$. Rewrite: $$2x^2 - 16x - x + 8 < 0$$ Group terms: $$(2x^2 - 16x) - (x - 8) < 0$$ Factor each group: $$2x(x - 8) - 1(x - 8) < 0$$ Factor out $$(x - 8)$$: $$(2x - 1)(x - 8) < 0$$ 7. **Find critical points:** Set each factor equal to zero: $$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$$ $$x - 8 = 0 \Rightarrow x = 8$$ 8. **Determine intervals to test:** The critical points divide the number line into three intervals: $$(-\infty, \frac{1}{2}), (\frac{1}{2}, 8), (8, \infty)$$ 9. **Test each interval:** - For $$x = 0$$ in $$(-\infty, \frac{1}{2})$$: $$(2(0) - 1)(0 - 8) = (-1)(-8) = 8 > 0$$ (Not less than 0) - For $$x = 1$$ in $$(\frac{1}{2}, 8)$$: $$(2(1) - 1)(1 - 8) = (2 - 1)(-7) = (1)(-7) = -7 < 0$$ (Satisfies inequality) - For $$x = 9$$ in $$(8, \infty)$$: $$(2(9) - 1)(9 - 8) = (18 - 1)(1) = 17 > 0$$ (Not less than 0) 10. **Conclusion:** The solution to the inequality is $$\boxed{\left(\frac{1}{2}, 8\right)}$$ This means all $$x$$ values strictly between $$\frac{1}{2}$$ and $$8$$ satisfy the original inequality.