I. ` x^(2) + 28x + 192 = 0`
II. ` y^(2)+16y+48=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 and find the relationship between the values of \( x \) and \( y \), we will follow these steps:

### Step 1: Solve the first equation \( x^2 + 28x + 192 = 0 \)

1. **Identify the coefficients**: Here, \( a = 1 \), \( b = 28 \), and \( c = 192 \).
2. **Find factors of \( c \) that add up to \( b \)**: We need two numbers that multiply to \( 192 \) and add up to \( 28 \). The factors \( 16 \) and \( 12 \) satisfy this condition because:
- \( 16 \times 12 = 192 \)
- \( 16 + 12 = 28 \)
3. **Write the equation in factored form**:
\[ (x + 16)(x + 12) = 0 \]
4. **Set each factor to zero**:
\[ x + 16 = 0 \quad \Rightarrow \quad x = -16 \]
\[ x + 12 = 0 \quad \Rightarrow \quad x = -12 \]

### Step 2: Solve the second equation \( y^2 + 16y + 48 = 0 \)

1. **Identify the coefficients**: Here, \( a = 1 \), \( b = 16 \), and \( c = 48 \).
2. **Find factors of \( c \) that add up to \( b \)**: We need two numbers that multiply to \( 48 \) and add up to \( 16 \). The factors \( 12 \) and \( 4 \) satisfy this condition because:
- \( 12 \times 4 = 48 \)
- \( 12 + 4 = 16 \)
3. **Write the equation in factored form**:
\[ (y + 12)(y + 4) = 0 \]
4. **Set each factor to zero**:
\[ y + 12 = 0 \quad \Rightarrow \quad y = -12 \]
\[ y + 4 = 0 \quad \Rightarrow \quad y = -4 \]

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

- The values of \( x \) are \( -16 \) and \( -12 \).
- The values of \( y \) are \( -12 \) and \( -4 \).

### Step 4: Determine the relationship between \( x \) and \( y \)

- Compare \( x = -12 \) with \( y = -12 \): They are equal.
- Compare \( x = -16 \) with \( y = -4 \): Here, \( -16 < -4 \).

Thus, we can conclude:
- \( x \) can be equal to \( y \) when both are \( -12 \).
- \( x \) is less than \( y \) when \( x = -16 \) and \( y = -4 \).

### Final Conclusion
The relationship between \( x \) and \( y \) is:
\[ x \leq y \]