I. ` 20x^(2) - x - 12 = 0`
II. ` 20y^(2) + 27y + 9 = 0`

Options

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

Explanation

To solve the given equations step by step, we will first tackle each quadratic equation separately.

### Step 1: Solve the first equation `20x^2 - x - 12 = 0`

1. **Identify the coefficients**:
- Here, \( a = 20 \), \( b = -1 \), and \( c = -12 \).

2. **Factor the quadratic**:
- We need to find two numbers that multiply to \( ac = 20 \times -12 = -240 \) and add to \( b = -1 \).
- The numbers that satisfy this are \( 15 \) and \( -16 \) because \( 15 \times -16 = -240 \) and \( 15 + (-16) = -1 \).

3. **Rewrite the equation**:
- We can rewrite the equation as:
\[ 20x^2 + 15x - 16x - 12 = 0 \]

4. **Group the terms**:
- Group the terms:
\[ (20x^2 + 15x) + (-16x - 12) = 0 \]

5. **Factor by grouping**:
- Factor out common terms:
\[ 5x(4x + 3) - 4(4x + 3) = 0 \]
- This gives us:
\[ (5x - 4)(4x + 3) = 0 \]

6. **Set each factor to zero**:
- Solve for \( x \):
\[ 5x - 4 = 0 \quad \Rightarrow \quad x = \frac{4}{5} \]
\[ 4x + 3 = 0 \quad \Rightarrow \quad x = -\frac{3}{4} \]

### Step 2: Solve the second equation `20y^2 + 27y + 9 = 0`

1. **Identify the coefficients**:
- Here, \( a = 20 \), \( b = 27 \), and \( c = 9 \).

2. **Factor the quadratic**:
- We need to find two numbers that multiply to \( ac = 20 \times 9 = 180 \) and add to \( b = 27 \).
- The numbers that satisfy this are \( 15 \) and \( 12 \) because \( 15 \times 12 = 180 \) and \( 15 + 12 = 27 \).

3. **Rewrite the equation**:
- We can rewrite the equation as:
\[ 20y^2 + 15y + 12y + 9 = 0 \]

4. **Group the terms**:
- Group the terms:
\[ (20y^2 + 15y) + (12y + 9) = 0 \]

5. **Factor by grouping**:
- Factor out common terms:
\[ 5y(4y + 3) + 3(4y + 3) = 0 \]
- This gives us:
\[ (5y + 3)(4y + 3) = 0 \]

6. **Set each factor to zero**:
- Solve for \( y \):
\[ 5y + 3 = 0 \quad \Rightarrow \quad y = -\frac{3}{5} \]
\[ 4y + 3 = 0 \quad \Rightarrow \quad y = -\frac{3}{4} \]

### Summary of Solutions:
- The solutions for \( x \) are \( \frac{4}{5} \) and \( -\frac{3}{4} \).
- The solutions for \( y \) are \( -\frac{3}{5} \) and \( -\frac{3}{4} \).