Let's solve each of these quadratic equations using the extracting square roots method, where applicable.

1. $x^2 = 169$

To solve: $x = \pm \sqrt{169}$ $x = \pm 13$

Solution: $x = 13$ or $x = -13$

2. $x^2 + 3 = 39$

First, isolate $x^2$: $x^2 = 39 - 3$ $x^2 = 36$

Now, take the square root: $x = \pm \sqrt{36}$ $x = \pm 6$

Solution: $x = 6$ or $x = -6$

3. $5x^2 - 7 = 118$

Isolate $x^2$: $5x^2 = 118 + 7$ $5x^2 = 125$ $x^2 = \frac{125}{5}$ $x^2 = 25$

Take the square root: $x = \pm \sqrt{25}$ $x = \pm 5$

Solution: $x = 5$ or $x = -5$

4. $x^2 + 103 = 3$

Isolate $x^2$: $x^2 = 3 - 103$ $x^2 = -100$

Since $x^2 = -100$ has no real solutions (square roots of negative numbers are not real), this equation has no real solutions.

Solution: No real solutions.

5. $4x^2 - 32 = 0$

Isolate $x^2$: $4x^2 = 32$ $x^2 = \frac{32}{4}$ $x^2 = 8$

Now, take the square root: $x = \pm \sqrt{8}$ $x = \pm 2\sqrt{2}$

Solution: $x = 2\sqrt{2}$ or $x = -2\sqrt{2}$

Let me know if you have any questions about these solutions or need further clarification!

Here are some related questions:

1. How would you solve these equations using factoring instead of square roots?
2. What are some real-world applications of quadratic equations?
3. How would these solutions change if we were looking for complex roots?
4. Why does $x^2 = -100$ have no real solution?
5. Can these methods be extended to equations of higher degree, like $x^4$?

Tip: Always check if the equation can be simplified before taking square roots. This can make the solving process faster and more accurate!