In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
I `2x^2+5x+2=0`
II `y^2-12y+20=0`

Options

A. `xgty`
B. `xgey`
C. `xlty`
D. `xley`

Answer

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

### Step 1: Solve the first equation \(2x^2 + 5x + 2 = 0\)

1. **Identify the coefficients**: Here, \(a = 2\), \(b = 5\), and \(c = 2\).
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 into the formula**:
\[ x = \frac{-5 \pm \sqrt{9}}{2 \cdot 2} = \frac{-5 \pm 3}{4} \]
5. **Calculate the two possible values for \(x\)**:
- For \(x_1\):
\[ x_1 = \frac{-5 + 3}{4} = \frac{-2}{4} = -\frac{1}{2} \]
- For \(x_2\):
\[ x_2 = \frac{-5 - 3}{4} = \frac{-8}{4} = -2 \]

Thus, the solutions for the first equation are:
\[ x = -\frac{1}{2}, \quad x = -2 \]

### Step 2: Solve the second equation \(y^2 - 12y + 20 = 0\)

1. **Identify the coefficients**: Here, \(a = 1\), \(b = -12\), and \(c = 20\).
2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-12)^2 - 4 \cdot 1 \cdot 20 = 144 - 80 = 64 \]
4. **Substitute into the formula**:
\[ y = \frac{12 \pm \sqrt{64}}{2 \cdot 1} = \frac{12 \pm 8}{2} \]
5. **Calculate the two possible values for \(y\)**:
- For \(y_1\):
\[ y_1 = \frac{12 + 8}{2} = \frac{20}{2} = 10 \]
- For \(y_2\):
\[ y_2 = \frac{12 - 8}{2} = \frac{4}{2} = 2 \]

Thus, the solutions for the second equation are:
\[ y = 10, \quad y = 2 \]

### Summary of Solutions:
- From the first equation: \(x = -\frac{1}{2}, -2\)
- From the second equation: \(y = 10, 2\)

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

Now we compare the values of \(x\) and \(y\):
- The maximum value of \(y\) is \(10\) and the minimum value of \(x\) is \(-2\).
- Therefore, \(y\) is always greater than \(x\).

### Final Answer:
The relationship is:
\[ x < y \]