Let's solve these equations one by one. Each involves solving for $x$ where the right-hand side is not a perfect square. Here's how we approach them:

1. $(x + 3)^2 = 7$

1. First, take the square root of both sides: $x + 3 = \pm \sqrt{7}$

2. Now, solve for $x$: $x = -3 \pm \sqrt{7}$

So, the two possible solutions for this equation are: $x = -3 + \sqrt{7} \quad \text{or} \quad x = -3 - \sqrt{7}$

2. $(x - 2)^2 = 11$

1. Take the square root of both sides: $x - 2 = \pm \sqrt{11}$

2. Solve for $x$: $x = 2 \pm \sqrt{11}$

Thus, the solutions are: $x = 2 + \sqrt{11} \quad \text{or} \quad x = 2 - \sqrt{11}$

3. $(x + 4)^2 = 18$

1. Take the square root of both sides: $x + 4 = \pm \sqrt{18}$

2. Simplify $\sqrt{18}$ to $3\sqrt{2}$: $x + 4 = \pm 3\sqrt{2}$

3. Solve for $x$: $x = -4 \pm 3\sqrt{2}$

So, the solutions are: $x = -4 + 3\sqrt{2} \quad \text{or} \quad x = -4 - 3\sqrt{2}$

4. $(x - 5)^2 = 20$

1. Take the square root of both sides: $x - 5 = \pm \sqrt{20}$

2. Simplify $\sqrt{20}$ to $2\sqrt{5}$: $x - 5 = \pm 2\sqrt{5}$

3. Solve for $x$: $x = 5 \pm 2\sqrt{5}$

The solutions are: $x = 5 + 2\sqrt{5} \quad \text{or} \quad x = 5 - 2\sqrt{5}$

5. $(x + 2)^2 = 13$

1. Take the square root of both sides: $x + 2 = \pm \sqrt{13}$

2. Solve for $x$: $x = -2 \pm \sqrt{13}$

So, the solutions are: $x = -2 + \sqrt{13} \quad \text{or} \quad x = -2 - \sqrt{13}$

Let me know if you'd like to proceed with the remaining problems or need additional clarifications!