1. **State the problem:** Solve the inequality $$x(x+2) - 3(2x - 1) \leq 2^2$$. 2. **Rewrite the inequality:** Expand and simplify the left side. $$x(x+2) - 3(2x - 1) \leq 4$$ $$x^2 + 2x - 6x + 3 \leq 4$$ 3. **Combine like terms:** $$x^2 - 4x + 3 \leq 4$$ 4. **Bring all terms to one side:** $$x^2 - 4x + 3 - 4 \leq 0$$ $$x^2 - 4x - 1 \leq 0$$ 5. **Solve the quadratic inequality:** First, find the roots of the quadratic equation $$x^2 - 4x - 1 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-4$$, and $$c=-1$$. Calculate the discriminant: $$\Delta = (-4)^2 - 4(1)(-1) = 16 + 4 = 20$$ Calculate the roots: $$x = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$$ 6. **Determine the solution set:** Since the parabola opens upwards (coefficient of $$x^2$$ is positive), the inequality $$x^2 - 4x - 1 \leq 0$$ holds between the roots. Therefore, the solution is: $$2 - \sqrt{5} \leq x \leq 2 + \sqrt{5}$$ **Final answer:** $$\boxed{2 - \sqrt{5} \leq x \leq 2 + \sqrt{5}}$$