In each of these questions, two equations I. and II. are given. You have to solve both the equations and give answer
I. ` 2 x^(2) - 3 x - 20 = 0 `
II. ` 2y^(2) + 11 y + 15 = 0 `

Options

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

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

### Step 1: Solve Equation I
The first equation is:
\[ 2x^2 - 3x - 20 = 0 \]

#### Step 1.1: Factor the equation
To factor the quadratic equation, we need to find two numbers that multiply to \(2 \times -20 = -40\) and add up to \(-3\). The numbers that satisfy this are \(5\) and \(-8\).

We can rewrite the equation as:
\[ 2x^2 - 8x + 5x - 20 = 0 \]

#### Step 1.2: Group the terms
Now, we group the terms:
\[ (2x^2 - 8x) + (5x - 20) = 0 \]

#### Step 1.3: Factor by grouping
Factoring out the common factors, we get:
\[ 2x(x - 4) + 5(x - 4) = 0 \]

Now, we can factor out \((x - 4)\):
\[ (x - 4)(2x + 5) = 0 \]

#### Step 1.4: Solve for \(x\)
Setting each factor to zero gives us:
1. \(x - 4 = 0 \Rightarrow x = 4\)
2. \(2x + 5 = 0 \Rightarrow 2x = -5 \Rightarrow x = -\frac{5}{2} = -2.5\)

So, the solutions for \(x\) are:
\[ x = 4 \quad \text{and} \quad x = -2.5 \]

### Step 2: Solve Equation II
The second equation is:
\[ 2y^2 + 11y + 15 = 0 \]

#### Step 2.1: Factor the equation
We need to find two numbers that multiply to \(2 \times 15 = 30\) and add up to \(11\). The numbers that satisfy this are \(5\) and \(6\).

We can rewrite the equation as:
\[ 2y^2 + 6y + 5y + 15 = 0 \]

#### Step 2.2: Group the terms
Now, we group the terms:
\[ (2y^2 + 6y) + (5y + 15) = 0 \]

#### Step 2.3: Factor by grouping
Factoring out the common factors, we get:
\[ 2y(y + 3) + 5(y + 3) = 0 \]

Now, we can factor out \((y + 3)\):
\[ (y + 3)(2y + 5) = 0 \]

#### Step 2.4: Solve for \(y\)
Setting each factor to zero gives us:
1. \(y + 3 = 0 \Rightarrow y = -3\)
2. \(2y + 5 = 0 \Rightarrow 2y = -5 \Rightarrow y = -\frac{5}{2} = -2.5\)

So, the solutions for \(y\) are:
\[ y = -3 \quad \text{and} \quad y = -2.5 \]

### Step 3: Compare the values of \(x\) and \(y\)
We have:
- \(x = 4\) or \(x = -2.5\)
- \(y = -3\) or \(y = -2.5\)

#### Step 3.1: Compare \(x\) and \(y\)
1. For \(x = 4\) and \(y = -3\):
- \(4 > -3\)

2. For \(x = -2.5\) and \(y = -3\):
- \(-2.5 > -3\)

3. For \(x = -2.5\) and \(y = -2.5\):
- \(-2.5 = -2.5\)

### Conclusion
From the comparisons, we can conclude:
- \(x\) is greater than \(y\) in both cases where \(x = 4\) or \(x = -2.5\).

Thus, the final answer is:
\[ x \geq y \]