` 2x^(2) - 13x + 20 = 0, 2y^(2) - 7y + 6 = 0`

Options

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

Explanation

To solve the equations \( 2x^2 - 13x + 20 = 0 \) and \( 2y^2 - 7y + 6 = 0 \), we will follow these steps:

### Step 1: Solve for \( x \) in the equation \( 2x^2 - 13x + 20 = 0 \)

1. **Identify the coefficients**: Here, \( a = 2 \), \( b = -13 \), and \( c = 20 \).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-13)^2 - 4 \cdot 2 \cdot 20 = 169 - 160 = 9 \]
3. **Use the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{13 \pm \sqrt{9}}{4} = \frac{13 \pm 3}{4} \]
This gives us two possible values for \( x \):
\[ x_1 = \frac{16}{4} = 4, \quad x_2 = \frac{10}{4} = \frac{5}{2} \]

### Step 2: Solve for \( y \) in the equation \( 2y^2 - 7y + 6 = 0 \)

1. **Identify the coefficients**: Here, \( a = 2 \), \( b = -7 \), and \( c = 6 \).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot 6 = 49 - 48 = 1 \]
3. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{1}}{4} = \frac{7 \pm 1}{4} \]
This gives us two possible values for \( y \):
\[ y_1 = \frac{8}{4} = 2, \quad y_2 = \frac{6}{4} = \frac{3}{2} \]

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

1. **List the values obtained**:
- \( x_1 = 4 \)
- \( x_2 = \frac{5}{2} = 2.5 \)
- \( y_1 = 2 \)
- \( y_2 = \frac{3}{2} = 1.5 \)

2. **Order the values**:
- The values of \( x \) are \( 2.5 \) and \( 4 \).
- The values of \( y \) are \( 1.5 \) and \( 2 \).

3. **Determine the relationship**:
- Comparing \( x \) and \( y \):
- \( 4 > 2 \)
- \( 2.5 > 2 \)
- \( 2 > 1.5 \)

Thus, we conclude that \( x \) is greater than \( y \).

### Final Result
The relationship between \( x \) and \( y \) is:
\[ x > y \]

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