Solve the given quadratic equations and mark the correct options based on your answer
I. ` 7 x ^(2) + 5 x - 18 = 0 `
II. ` 3 y ^(2) + 4y - 20 = 0 `

Options

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

Answer

To solve the given quadratic equations, we will follow these steps:

### Step 1: Solve the first quadratic equation \(7x^2 + 5x - 18 = 0\)

1. **Identify coefficients**: Here, \(a = 7\), \(b = 5\), and \(c = -18\).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 5^2 - 4 \cdot 7 \cdot (-18) = 25 + 504 = 529 \]
3. **Find the roots using the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{529}}{2 \cdot 7} \]
Since \(\sqrt{529} = 23\), we have:
\[ x = \frac{-5 \pm 23}{14} \]
4. **Calculate the two possible values for \(x\)**:
- For \(x_1\):
\[ x_1 = \frac{-5 + 23}{14} = \frac{18}{14} = \frac{9}{7} \approx 1.2857 \]
- For \(x_2\):
\[ x_2 = \frac{-5 - 23}{14} = \frac{-28}{14} = -2 \]

### Step 2: Solve the second quadratic equation \(3y^2 + 4y - 20 = 0\)

1. **Identify coefficients**: Here, \(a = 3\), \(b = 4\), and \(c = -20\).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot (-20) = 16 + 240 = 256 \]
3. **Find the roots using the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{256}}{2 \cdot 3} \]
Since \(\sqrt{256} = 16\), we have:
\[ y = \frac{-4 \pm 16}{6} \]
4. **Calculate the two possible values for \(y\)**:
- For \(y_1\):
\[ y_1 = \frac{-4 + 16}{6} = \frac{12}{6} = 2 \]
- For \(y_2\):
\[ y_2 = \frac{-4 - 16}{6} = \frac{-20}{6} = -\frac{10}{3} \approx -3.3333 \]

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

We have the values:
- \(x_1 \approx 1.2857\), \(x_2 = -2\)
- \(y_1 = 2\), \(y_2 \approx -3.3333\)

Now, we will compare:
1. \(x_1 \approx 1.2857\) and \(y_1 = 2\): \(x_1 < y_1\)
2. \(x_1 \approx 1.2857\) and \(y_2 \approx -3.3333\): \(x_1 > y_2\)
3. \(x_2 = -2\) and \(y_1 = 2\): \(x_2 < y_1\)
4. \(x_2 = -2\) and \(y_2 \approx -3.3333\): \(x_2 > y_2\)

### Conclusion

From the comparisons:
- \(x_1 < y_1\)
- \(x_1 > y_2\)
- \(x_2 < y_1\)
- \(x_2 > y_2\)

This indicates that \(x\) can be both greater than and less than \(y\) depending on the specific values chosen. Therefore, no definitive relationship can be established between \(x\) and \(y\).

### Final Answer
The values of \(x\) and \(y\) are:
- \(x_1 \approx 1.2857\), \(x_2 = -2\)
- \(y_1 = 2\), \(y_2 \approx -3.3333\)