Directions:In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer.
I. `2x^2+5x+2=0` II. `4y^2-7y-2=0`

Options

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

Answer

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

### Step 1: Solve Equation I: \(2x^2 + 5x + 2 = 0\)

1. **Identify the coefficients**:
- \(a = 2\), \(b = 5\), \(c = 2\)

2. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \]

4. **Substitute values into the quadratic formula**:
\[ x = \frac{-5 \pm \sqrt{9}}{2 \cdot 2} = \frac{-5 \pm 3}{4} \]

5. **Calculate the two possible values for \(x\)**:
- First value:
\[ x_1 = \frac{-5 + 3}{4} = \frac{-2}{4} = -\frac{1}{2} \]
- Second value:
\[ x_2 = \frac{-5 - 3}{4} = \frac{-8}{4} = -2 \]

### Step 2: Solve Equation II: \(4y^2 - 7y - 2 = 0\)

1. **Identify the coefficients**:
- \(a = 4\), \(b = -7\), \(c = -2\)

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 4 \cdot (-2) = 49 + 32 = 81 \]

4. **Substitute values into the quadratic formula**:
\[ y = \frac{7 \pm \sqrt{81}}{2 \cdot 4} = \frac{7 \pm 9}{8} \]

5. **Calculate the two possible values for \(y\)**:
- First value:
\[ y_1 = \frac{7 + 9}{8} = \frac{16}{8} = 2 \]
- Second value:
\[ y_2 = \frac{7 - 9}{8} = \frac{-2}{8} = -\frac{1}{4} \]

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

- Possible values for \(x\) are \(-\frac{1}{2}\) and \(-2\).
- Possible values for \(y\) are \(2\) and \(-\frac{1}{4}\).

Now we compare:

1. For \(x = -\frac{1}{2}\) and \(y = -\frac{1}{4}\):
\[ -\frac{1}{2} < -\frac{1}{4} \quad \text{(True)} \]

2. For \(x = -2\) and \(y = -\frac{1}{4}\):
\[ -2 < -\frac{1}{4} \quad \text{(True)} \]

3. For \(x = -\frac{1}{2}\) and \(y = 2\):
\[ -\frac{1}{2} < 2 \quad \text{(True)} \]

4. For \(x = -2\) and \(y = 2\):
\[ -2 < 2 \quad \text{(True)} \]

### Conclusion
From the comparisons, we can conclude that in all cases, \(x < y\).

### Final Answer
The relation is \(x < y\).