In each question two equations numbered I and II are given. You have to solve both the equations and mark the answer
I. `7x^2 +12x+5=0` II. `3y^2 +7y +2=0`

Options

A. if `x gt y`
B. if `x ge y `
C. if `x lt y `
D. `if ` x=y `or no relation can be established between x and y.

Answer

To solve the given equations and find the relationship between the values of \( x \) and \( y \), we will follow these steps:

### Step 1: Solve the first equation \( 7x^2 + 12x + 5 = 0 \)

We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 7 \), \( b = 12 \), and \( c = 5 \).

Calculating the discriminant:
\[ b^2 - 4ac = 12^2 - 4 \cdot 7 \cdot 5 = 144 - 140 = 4 \]
Since the discriminant is positive, we have two real roots.

Now substituting into the quadratic formula:
\[ x = \frac{-12 \pm \sqrt{4}}{2 \cdot 7} = \frac{-12 \pm 2}{14} \]

Calculating the two possible values for \( x \):
1. \( x_1 = \frac{-12 + 2}{14} = \frac{-10}{14} = -\frac{5}{7} \)
2. \( x_2 = \frac{-12 - 2}{14} = \frac{-14}{14} = -1 \)

Thus, the solutions for \( x \) are:
\[ x = -\frac{5}{7} \quad \text{and} \quad x = -1 \]

### Step 2: Solve the second equation \( 3y^2 + 7y + 2 = 0 \)

Again, we will use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 3 \), \( b = 7 \), and \( c = 2 \).

Calculating the discriminant:
\[ b^2 - 4ac = 7^2 - 4 \cdot 3 \cdot 2 = 49 - 24 = 25 \]
Since the discriminant is positive, we have two real roots.

Now substituting into the quadratic formula:
\[ y = \frac{-7 \pm \sqrt{25}}{2 \cdot 3} = \frac{-7 \pm 5}{6} \]

Calculating the two possible values for \( y \):
1. \( y_1 = \frac{-7 + 5}{6} = \frac{-2}{6} = -\frac{1}{3} \)
2. \( y_2 = \frac{-7 - 5}{6} = \frac{-12}{6} = -2 \)

Thus, the solutions for \( y \) are:
\[ y = -\frac{1}{3} \quad \text{and} \quad y = -2 \]

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

We have the following values:
- \( x_1 = -\frac{5}{7} \approx -0.71 \)
- \( x_2 = -1 \)
- \( y_1 = -\frac{1}{3} \approx -0.33 \)
- \( y_2 = -2 \)

Now we will compare:
1. Comparing \( x_1 = -\frac{5}{7} \) and \( y_1 = -\frac{1}{3} \):
- Since \( -0.71 < -0.33 \), we have \( x_1 < y_1 \).

2. Comparing \( x_1 = -\frac{5}{7} \) and \( y_2 = -2 \):
- Since \( -0.71 > -2 \), we have \( x_1 > y_2 \).

3. Comparing \( x_2 = -1 \) and \( y_1 = -\frac{1}{3} \):
- Since \( -1 < -0.33 \), we have \( x_2 < y_1 \).

4. Comparing \( x_2 = -1 \) and \( y_2 = -2 \):
- Since \( -1 > -2 \), we have \( x_2 > y_2 \).

### Conclusion

From our comparisons, we see that:
- \( x_1 < y_1 \) and \( x_1 > y_2 \)
- \( x_2 < y_1 \) and \( x_2 > y_2 \)

Thus, we conclude that there is no consistent relationship between \( x \) and \( y \) across all cases.

### Final Answer
The answer is: **No relation can be established between \( x \) and \( y \)**.

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