I.` 12x^(2) + 82 x + 140 = 0`
II. ` 16y^(2) + 48y+ 32 = 0`

Options

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

Explanation

To solve the equations step by step, we will tackle each equation one by one.

### Step 1: Solve the first equation \( 12x^2 + 82x + 140 = 0 \)

1. **Divide the entire equation by 2** (to simplify):
\[ 6x^2 + 41x + 70 = 0 \]

2. **Find the product of the coefficient of \( x^2 \) and the constant term**:
\[ 6 \times 70 = 420 \]

3. **We need to find two numbers that multiply to 420 and add up to 41**. After checking various factor pairs, we find:
- The numbers are 20 and 21.

4. **Rewrite the middle term using these numbers**:
\[ 6x^2 + 20x + 21x + 70 = 0 \]

5. **Factor by grouping**:
- Group the first two terms and the last two terms:
\[ (6x^2 + 20x) + (21x + 70) = 0 \]
- Factor out common terms:
\[ 2x(3x + 10) + 7(3x + 10) = 0 \]
- Combine the factors:
\[ (2x + 7)(3x + 10) = 0 \]

6. **Set each factor to zero**:
\[ 2x + 7 = 0 \quad \text{or} \quad 3x + 10 = 0 \]

7. **Solve for \( x \)**:
- From \( 2x + 7 = 0 \):
\[ 2x = -7 \quad \Rightarrow \quad x = -\frac{7}{2} = -3.5 \]
- From \( 3x + 10 = 0 \):
\[ 3x = -10 \quad \Rightarrow \quad x = -\frac{10}{3} \approx -3.33 \]

### Step 2: Solve the second equation \( 16y^2 + 48y + 32 = 0 \)

1. **Divide the entire equation by 16**:
\[ y^2 + 3y + 2 = 0 \]

2. **Find the product of the coefficient of \( y^2 \) and the constant term**:
\[ 1 \times 2 = 2 \]

3. **We need to find two numbers that multiply to 2 and add up to 3**:
- The numbers are 1 and 2.

4. **Rewrite the middle term using these numbers**:
\[ y^2 + 1y + 2y + 2 = 0 \]

5. **Factor by grouping**:
- Group the first two terms and the last two terms:
\[ (y^2 + 1y) + (2y + 2) = 0 \]
- Factor out common terms:
\[ y(y + 1) + 2(y + 1) = 0 \]
- Combine the factors:
\[ (y + 1)(y + 2) = 0 \]

6. **Set each factor to zero**:
\[ y + 1 = 0 \quad \text{or} \quad y + 2 = 0 \]

7. **Solve for \( y \)**:
- From \( y + 1 = 0 \):
\[ y = -1 \]
- From \( y + 2 = 0 \):
\[ y = -2 \]

### Summary of Solutions:
- For \( x \): \( x = -\frac{10}{3} \approx -3.33 \) and \( x = -\frac{7}{2} = -3.5 \)
- For \( y \): \( y = -1 \) and \( y = -2 \)

### Step 3: Establish the relationship between \( x \) and \( y \)

- The values of \( x \) are approximately \( -3.33 \) and \( -3.5 \).
- The values of \( y \) are \( -1 \) and \( -2 \).

### Comparison:
- Since \( -3.5 < -2 < -1 < -3.33 \), we can conclude that:
\[ x < y \]