Solve the equation: 2) $$(x+2)^2 - 4(x+2) - 5 = 0$$

Let's solve the equation step by step: 1. We have the equation $$(x+2)^2 - 4(x+2) - 5 = 0$$. To simplify, let's make a substitution: $$t = x+2$$. Then the equation becomes: $$t^2 - 4t - 5 = 0$$. 2. Now, let's solve the quadratic equation $$t^2 - 4t - 5 = 0$$. We can factor this quadratic equation as: $$(t - 5)(t + 1) = 0$$. 3. This gives us two possible values for $$t$$: $$t - 5 = 0 \Rightarrow t_1 = 5$$ $$t + 1 = 0 \Rightarrow t_2 = -1$$ 4. Now, let's find the corresponding values for $$x$$ using the substitution $$t = x+2$$: For $$t_1 = 5$$: $$x+2 = 5 \Rightarrow x_1 = 5-2 = 3$$ For $$t_2 = -1$$: $$x+2 = -1 \Rightarrow x_2 = -1-2 = -3$$ Thus, the solutions for the equation are $$x_1 = 3$$ and $$x_2 = -3$$. Answer: $$x_1 = 3, x_2 = -3$$