In each of the following questions two equations are given. Solve these equations and give answer
I. ` 6 x^(2) + 17 x + 5 = 0 `
II. ` 2 y^(2) + 21 y + 49 = 0 `

Options

A. If ` x ge y ` , i.e. x is greater than or equal to y
B. If ` x gt y `, i.e. x is greater than y
C. If ` x le y`, i.e. x is less than or equal to y
D. If ` x lt y ` , i.e. x is less than y

Answer

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

### Step 1: Solve the first equation \( 6x^2 + 17x + 5 = 0 \)

1. **Identify the coefficients**:
- \( a = 6 \)
- \( b = 17 \)
- \( c = 5 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 17^2 - 4 \cdot 6 \cdot 5 = 289 - 120 = 169 \]

4. **Calculate the roots**:
\[ x = \frac{-17 \pm \sqrt{169}}{2 \cdot 6} = \frac{-17 \pm 13}{12} \]
- First root:
\[ x_1 = \frac{-17 + 13}{12} = \frac{-4}{12} = -\frac{1}{3} \]
- Second root:
\[ x_2 = \frac{-17 - 13}{12} = \frac{-30}{12} = -\frac{5}{2} \]

### Step 2: Solve the second equation \( 2y^2 + 21y + 49 = 0 \)

1. **Identify the coefficients**:
- \( a = 2 \)
- \( b = 21 \)
- \( c = 49 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 21^2 - 4 \cdot 2 \cdot 49 = 441 - 392 = 49 \]

4. **Calculate the roots**:
\[ y = \frac{-21 \pm \sqrt{49}}{2 \cdot 2} = \frac{-21 \pm 7}{4} \]
- First root:
\[ y_1 = \frac{-21 + 7}{4} = \frac{-14}{4} = -\frac{7}{2} \]
- Second root:
\[ y_2 = \frac{-21 - 7}{4} = \frac{-28}{4} = -7 \]

### Summary of Solutions
- The solutions for \( x \) are \( x_1 = -\frac{1}{3} \) and \( x_2 = -\frac{5}{2} \).
- The solutions for \( y \) are \( y_1 = -\frac{7}{2} \) and \( y_2 = -7 \).