Thank you for the clarification! Let's solve the quadratic equations listed in the image.

### 1. \( x^2 + 16 = 4 \)
Start by moving 4 to the left side:
\[
x^2 + 16 - 4 = 0
\]
\[
x^2 + 12 = 0
\]
Subtract 12 from both sides:
\[
x^2 = -12
\]
Take the square root of both sides:
\[
x = \pm \sqrt{-12} = \pm 2i\sqrt{3}
\]
(Since the square root of a negative number introduces an imaginary unit \(i\)).

### 2. \( x^2 + 9 = 7 \)
Start by moving 7 to the left side:
\[
x^2 + 9 - 7 = 0
\]
\[
x^2 + 2 = 0
\]
Subtract 2 from both sides:
\[
x^2 = -2
\]
Take the square root of both sides:
\[
x = \pm \sqrt{-2} = \pm i\sqrt{2}
\]
(This also has imaginary solutions).

### 3. \( x^2 - 80 = 0 \)
Add 80 to both sides:
\[
x^2 = 80
\]
Take the square root of both sides:
\[
x = \pm \sqrt{80} = \pm 4\sqrt{5}
\]

### 4. \( x^2 + 132 = x \)
Move \( x \) to the left side to form a standard quadratic equation:
\[
x^2 - x + 132 = 0
\]
Now, use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), and \(c = 132\):
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(132)}}{2(1)}
\]
\[
x = \frac{1 \pm \sqrt{1 - 528}}{2}
\]
\[
x = \frac{1 \pm \sqrt{-527}}{2} = \frac{1 \pm i\sqrt{527}}{2}
\]
So the solutions are complex: \(x = \frac{1 + i\sqrt{527}}{2}\) and \(x = \frac{1 - i\sqrt{527}}{2}\).

### 5. \( x^2 + 1 = 0 \)
Subtract 1 from both sides:
\[
x^2 = -1
\]
Take the square root of both sides:
\[
x = \pm \sqrt{-1} = \pm i
\]

### 6. \( x^2 - 65 = 0 \)
Add 65 to both sides:
\[
x^2 = 65
\]
Take the square root of both sides:
\[
x = \pm \sqrt{65}
\]

### 7. \( x^2 - 6 = x + 7 \)
Move all terms to one side:
\[
x^2 - x - 6 - 7 = 0
\]
\[
x^2 - x - 13 = 0
\]
Now use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), and \(c = -13\):
\[
x = \frac{1 \pm \sqrt{1 + 52}}{2} = \frac{1 \pm \sqrt{53}}{2}
\]

### 8. \( x^2 + 1.25 = 0 \)
Subtract 1.25 from both sides:
\[
x^2 = -1.25
\]
Take the square root:
\[
x = \pm \sqrt{-1.25} = \pm \sqrt{\frac{-5}{4}} = \pm \frac{\sqrt{5}i}{2}
\]

### 9. **(Repeated from above):** \( x^2 + 1.25 = 0 \)
The solution remains \( x = \pm \frac{\sqrt{5}i}{2} \).

If you'd like to see more details or have further questions, let me know!

---

### Additional Questions:
1. How do you solve quadratic equations using the quadratic formula?
2. What are the implications of having a negative discriminant in a quadratic equation?
3. Can you explain the process of completing the square for solving quadratic equations?
4. How do you graph the solutions of quadratic equations?
5. What does it mean if a quadratic equation has no real solutions?
6. How do the solutions of a quadratic equation relate to the graph of the corresponding parabola?
7. How can you factor a quadratic equation if it is factorable?
8. What are the possible types of roots a quadratic equation can have?

### Tip:
Always check the discriminant \((b^2 - 4ac)\) before solving a quadratic equation to determine the nature of the roots (real and distinct, real and equal, or complex).

Explore how to solve quadratic equations step-by-step with detailed examples. Learn about the quadratic formula and its application in finding solutions to equations involving square terms. Suitable for high school students and those preparing for standardized tests.

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