I. ` x^(2)+8x+15 = 0`
II. ` y^(2) + 11y + 30 = 0`

Options

A. `x gt y `
B. `x ge y`
C. ` x lt y`
D. ` x le y`

Explanation

To solve the given quadratic equations step by step, we will start with the first equation and then move on to the second one.

### Step 1: Solve the first equation \( x^2 + 8x + 15 = 0 \)

1. **Identify the coefficients**: Here, \( a = 1 \), \( b = 8 \), and \( c = 15 \).
2. **Factor the quadratic**: We need to find two numbers that multiply to \( 15 \) (the constant term) and add up to \( 8 \) (the coefficient of \( x \)). The numbers are \( 5 \) and \( 3 \).
3. **Write the factors**: The equation can be factored as:
\[ (x + 5)(x + 3) = 0 \]
4. **Set each factor to zero**:
\[ x + 5 = 0 \quad \text{or} \quad x + 3 = 0 \]
5. **Solve for \( x \)**:
- From \( x + 5 = 0 \), we get \( x = -5 \).
- From \( x + 3 = 0 \), we get \( x = -3 \).

### Step 2: Solve the second equation \( y^2 + 11y + 30 = 0 \)

1. **Identify the coefficients**: Here, \( a = 1 \), \( b = 11 \), and \( c = 30 \).
2. **Factor the quadratic**: We need to find two numbers that multiply to \( 30 \) (the constant term) and add up to \( 11 \) (the coefficient of \( y \)). The numbers are \( 6 \) and \( 5 \).
3. **Write the factors**: The equation can be factored as:
\[ (y + 6)(y + 5) = 0 \]
4. **Set each factor to zero**:
\[ y + 6 = 0 \quad \text{or} \quad y + 5 = 0 \]
5. **Solve for \( y \)**:
- From \( y + 6 = 0 \), we get \( y = -6 \).
- From \( y + 5 = 0 \), we get \( y = -5 \).

### Summary of Solutions

- The values of \( x \) are \( -5 \) and \( -3 \).
- The values of \( y \) are \( -6 \) and \( -5 \).

### Step 3: Compare values of \( x \) and \( y \)

- The pairs of solutions are:
- \( (x = -5, y = -5) \)
- \( (x = -3, y = -6) \)

From the pairs, we can see that:
- In the first pair, \( x \) and \( y \) are equal.
- In the second pair, \( x = -3 \) is greater than \( y = -6 \).

### Conclusion

Thus, we can conclude that \( x \) is greater than or equal to \( y \).