1. **State the problem:** Solve the equation $$|x^2 - 2x - 16| = 8$$ and verify solutions graphically. 2. **Recall the definition of absolute value:** For any expression $A$, $$|A| = B$$ means either $$A = B$$ or $$A = -B$$, provided $B \geq 0$. 3. **Apply this to our equation:** Let $$A = x^2 - 2x - 16$$ and $$B = 8$$. So, we have two cases: **Case 1:** $$x^2 - 2x - 16 = 8$$ **Case 2:** $$x^2 - 2x - 16 = -8$$ 4. **Solve Case 1:** $$x^2 - 2x - 16 = 8$$ Move all terms to one side: $$x^2 - 2x - 16 - 8 = 0$$ $$x^2 - 2x - 24 = 0$$ Factor or use quadratic formula: Discriminant $$D = (-2)^2 - 4(1)(-24) = 4 + 96 = 100$$ Solutions: $$x = \frac{2 \pm \sqrt{100}}{2} = \frac{2 \pm 10}{2}$$ So, $$x = \frac{2 + 10}{2} = 6$$ $$x = \frac{2 - 10}{2} = -4$$ 5. **Solve Case 2:** $$x^2 - 2x - 16 = -8$$ Move all terms to one side: $$x^2 - 2x - 16 + 8 = 0$$ $$x^2 - 2x - 8 = 0$$ Discriminant: $$D = (-2)^2 - 4(1)(-8) = 4 + 32 = 36$$ Solutions: $$x = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}$$ So, $$x = \frac{2 + 6}{2} = 4$$ $$x = \frac{2 - 6}{2} = -2$$ 6. **Verify solutions graphically:** The equation $$|x^2 - 2x - 16| = 8$$ means the graphs of $$y = |x^2 - 2x - 16|$$ and $$y = 8$$ intersect at the solutions. The solutions found are $$x = -4, -2, 4, 6$$. 7. **Final answer:** $$\boxed{x = -4, -2, 4, 6}$$