Let's match each equation with its solution based on the image you've uploaded. We will go step by step through each equation.

1. $x^2 = 25$

This is a straightforward quadratic equation. Taking the square root of both sides: $x = \pm \sqrt{25} = \pm 5$ Solution: $\pm 5$

2. $x^2 = -25$

Taking the square root of both sides: $x = \pm \sqrt{-25} = \pm 5i$ This involves imaginary numbers, and the solutions are $\pm 5i$.

Solution: $\pm 5i$

3. $x^2 + 5 = 0$

First, solve for $x^2$ by subtracting 5 from both sides: $x^2 = -5$ Now take the square root of both sides: $x = \pm \sqrt{-5} = \pm i\sqrt{5}$ The solutions are $\pm i\sqrt{5}$.

Solution: $\pm i\sqrt{5}$

4. $x^2 - 5 = 0$

First, solve for $x^2$ by adding 5 to both sides: $x^2 = 5$ Now take the square root of both sides: $x = \pm \sqrt{5}$ The solutions are $\pm \sqrt{5}$.

Solution: $\pm \sqrt{5}$

5. $x^2 = -45$

Taking the square root of both sides: $x = \pm \sqrt{-45} = \pm 3i\sqrt{5}$ This involves imaginary numbers. The solutions are $\pm 3i\sqrt{5}$.

Solution: $\pm 3i\sqrt{5}$

6. $x^2 = 45$

Taking the square root of both sides: $x = \pm \sqrt{45} = \pm 3\sqrt{5}$ The solutions are $\pm 3\sqrt{5}$.

Solution: $\pm 3\sqrt{5}$

7. $x - 5 = 0$

Solving for $x$ directly: $x = 5$ The solution is $5$.

Solution: $5$

8. $x + 5 = 0$

Solving for $x$ directly: $x = -5$ The solution is $-5$.

Solution: $-5$

Final Matching:

1. $x^2 = 25$ → $\pm 5$
2. $x^2 = -25$ → $\pm 5i$
3. $x^2 + 5 = 0$ → $\pm i\sqrt{5}$
4. $x^2 - 5 = 0$ → $\pm \sqrt{5}$
5. $x^2 = -45$ → $\pm 3i\sqrt{5}$
6. $x^2 = 45$ → $\pm 3\sqrt{5}$
7. $x - 5 = 0$ → $5$
8. $x + 5 = 0$ → $-5$

Do you have any questions or would you like more details?

Here are some related questions to deepen your understanding:

1. How do we solve quadratic equations with complex numbers?
2. Why do square roots of negative numbers involve imaginary numbers?
3. Can every quadratic equation have real solutions?
4. How can we apply the quadratic formula to solve these types of equations?
5. What is the difference between real and complex solutions in quadratic equations?

Tip: Always simplify square roots when possible, especially for negative numbers where imaginary solutions appear.