Solve the following quadratic equation and mark and answer as per instructions.
I. ` 2 x ^(2) + 5 x + 3 =0`
II. ` y^(2) + 4 y - 12 = 0`

Options

A. ` x le y`
B. ` x gt y`
C. x = y or no relation can be established
D. ` x lt y`

Answer

To solve the quadratic equations provided, we will follow a systematic approach for each equation.

### I. Solve the equation: \( 2x^2 + 5x + 3 = 0 \)

1. **Identify coefficients**:
- Here, \( a = 2 \), \( b = 5 \), and \( c = 3 \).

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot 3 = 25 - 24 = 1 \]

3. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values:
\[ x = \frac{-5 \pm \sqrt{1}}{2 \cdot 2} = \frac{-5 \pm 1}{4} \]

4. **Calculate the two possible values of \( x \)**:
- First value:
\[ x_1 = \frac{-5 + 1}{4} = \frac{-4}{4} = -1 \]
- Second value:
\[ x_2 = \frac{-5 - 1}{4} = \frac{-6}{4} = -1.5 \]

Thus, the solutions for the first equation are:
\[ x = -1 \quad \text{and} \quad x = -1.5 \]

### II. Solve the equation: \( y^2 + 4y - 12 = 0 \)

1. **Identify coefficients**:
- Here, \( a = 1 \), \( b = 4 \), and \( c = -12 \).

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-12) = 16 + 48 = 64 \]

3. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values:
\[ y = \frac{-4 \pm \sqrt{64}}{2 \cdot 1} = \frac{-4 \pm 8}{2} \]

4. **Calculate the two possible values of \( y \)**:
- First value:
\[ y_1 = \frac{-4 + 8}{2} = \frac{4}{2} = 2 \]
- Second value:
\[ y_2 = \frac{-4 - 8}{2} = \frac{-12}{2} = -6 \]

Thus, the solutions for the second equation are:
\[ y = 2 \quad \text{and} \quad y = -6 \]

### Summary of Solutions:
- For the first equation \( 2x^2 + 5x + 3 = 0 \):
- \( x = -1 \) and \( x = -1.5 \)
- For the second equation \( y^2 + 4y - 12 = 0 \):
- \( y = 2 \) and \( y = -6 \)

### Establishing Relation Between \( x \) and \( y \):
- Comparing the values:
- \( -1 \) is greater than \( -6 \) but less than \( 2 \).
- \( -1.5 \) is greater than \( -6 \) but less than \( 2 \).

From this comparison, we can conclude:
- \( x < y \) for both values of \( x \) when compared to \( y \).

### Final Answer:
The values of \( x \) are \( -1 \) and \( -1.5 \), and the values of \( y \) are \( 2 \) and \( -6 \). The relationship established is \( x < y \).