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Let's analyze the given quadratic equations to determine whether they have two real solutions, one real solution, or two imaginary solutions. This can be done by calculating the discriminant \( \Delta \) for each equation.

For a quadratic equation of the form:

\[
ax^2 + bx + c = 0
\]

The discriminant \( \Delta \) is given by:

\[
\Delta = b^2 - 4ac
\]

- If \( \Delta > 0 \), there are two distinct real solutions.
- If \( \Delta = 0 \), there is one real solution.
- If \( \Delta < 0 \), there are two imaginary (complex) solutions.

Let’s solve each equation step by step.

---

### 11. \( 3x^2 - 240 = 0 \)
Rewrite as: \( 3x^2 = 240 \)
This is a standard quadratic equation, and we can directly solve for the discriminant:
- \( a = 3 \)
- \( b = 0 \)
- \( c = -240 \)

Discriminant \( \Delta = 0^2 - 4(3)(-240) = 2880 \)

Since \( \Delta > 0 \), **two real solutions**.

---

### 12. \( -5x^2 = 60 \)
Rewrite as: \( -5x^2 - 60 = 0 \)
- \( a = -5 \)
- \( b = 0 \)
- \( c = -60 \)

Discriminant \( \Delta = 0^2 - 4(-5)(-60) = -1200 \)

Since \( \Delta < 0 \), **two imaginary solutions**.

---

### 13. \( \frac{x^2}{4} + 9 = 0 \)
Rewrite as: \( \frac{x^2}{4} = -9 \) or \( x^2 = -36 \)
- This gives imaginary solutions directly because \( x^2 = -36 \).

Thus, **two imaginary solutions**.

---

### 14. \( 7(x + 1)^2 - 24 = 0 \)
Rewrite as: \( 7(x + 1)^2 = 24 \), or \( (x + 1)^2 = \frac{24}{7} \)
- Taking the square root, we find two real roots.

Thus, **two real solutions**.

---

### 15. \( 2(3x^2 + 16) = 32 \)
Simplify to: \( 3x^2 + 16 = 16 \), then \( 3x^2 = 0 \)
- This simplifies to \( x = 0 \).

Thus, **one real solution**.

---

### 16. \( -\frac{x^2}{2} + 12 = 0 \)
Rewrite as: \( -\frac{x^2}{2} = -12 \), or \( x^2 = 24 \)
- This gives two real roots since \( x = \pm \sqrt{24} \).

Thus, **two real solutions**.

---

### Summary:
- 11: Two real solutions
- 12: Two imaginary solutions
- 13: Two imaginary solutions
- 14: Two real solutions
- 15: One real solution
- 16: Two real solutions

Would you like more details or have any questions about these calculations?

Here are five related questions:
1. What is the significance of the discriminant in quadratic equations?
2. How can you factor a quadratic equation to find its roots?
3. Can a quadratic equation have both real and imaginary solutions?
4. How do you solve a quadratic equation by completing the square?
5. What is the quadratic formula and when is it used?

**Tip**: When solving quadratic equations, always check if they can be factored easily before using the quadratic formula!

Determine if the following equations have two real solutions, one real solution, or two imaginary solutions.

Learn how to determine the number of real or imaginary solutions in quadratic equations using the discriminant. This guide provides step-by-step solutions for six example equations, suitable for students in Grades 9-12.

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