1. **State the problem:** Solve the inequality $$-2x^2 + 12x + 52 \leq -2$$ algebraically. 2. **Rewrite the inequality:** Move all terms to one side to set the inequality to zero: $$-2x^2 + 12x + 52 + 2 \leq 0$$ which simplifies to $$-2x^2 + 12x + 54 \leq 0$$ 3. **Divide the entire inequality by -2** to simplify, remembering to reverse the inequality sign because we divide by a negative number: $$\cancel{-2}x^2 \cancel{/\cancel{-2}} - \frac{12x}{-2} - \frac{54}{-2} \geq 0$$ which simplifies to $$x^2 - 6x - 27 \geq 0$$ 4. **Solve the quadratic equation** $$x^2 - 6x - 27 = 0$$ to find critical points. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-6$$, $$c=-27$$. Calculate the discriminant: $$\Delta = (-6)^2 - 4(1)(-27) = 36 + 108 = 144$$ Calculate roots: $$x = \frac{6 \pm \sqrt{144}}{2} = \frac{6 \pm 12}{2}$$ So, $$x_1 = \frac{6 - 12}{2} = \frac{-6}{2} = -3$$ $$x_2 = \frac{6 + 12}{2} = \frac{18}{2} = 9$$ 5. **Analyze the inequality** $$x^2 - 6x - 27 \geq 0$$. Since the parabola opens upward (coefficient of $$x^2$$ is positive), the expression is: - Greater than or equal to zero outside the roots - Less than zero between the roots 6. **Write the solution:** $$x \leq -3$$ or $$x \geq 9$$ **Final answer:** $$\boxed{x \leq -3 \text{ or } x \geq 9}$$