I.` 2x^(2) + 5x + 2 = 0" "II. Y^(2) + 9y + 20 = 0`

Options

A. if `x gt y`
B. if `x ge y`
C. if ` y gt x `
D. if ` y ge x`

Explanation

To solve the equations \(2x^2 + 5x + 2 = 0\) and \(y^2 + 9y + 20 = 0\), we will find the values of \(x\) and \(y\) step by step.

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

We can factor the quadratic equation. First, we rewrite it:

\[ 2x^2 + 4x + x + 2 = 0 \]

Now, we group the terms:

\[ (2x^2 + 4x) + (x + 2) = 0 \]

Factoring out the common terms:

\[ 2x(x + 2) + 1(x + 2) = 0 \]

Now, we can factor out \((x + 2)\):

\[ (x + 2)(2x + 1) = 0 \]

Setting each factor to zero gives us the possible values for \(x\):

1. \(x + 2 = 0 \implies x = -2\)
2. \(2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}\)

Thus, the solutions for \(x\) are:

\[ x = -2 \quad \text{and} \quad x = -\frac{1}{2} \]

### Step 2: Solve the second equation \(y^2 + 9y + 20 = 0\)

We can factor this quadratic equation as well. We look for two numbers that multiply to \(20\) and add up to \(9\). The numbers \(4\) and \(5\) work:

\[ y^2 + 4y + 5y + 20 = 0 \]

Grouping the terms:

\[ (y^2 + 4y) + (5y + 20) = 0 \]

Factoring out the common terms:

\[ y(y + 4) + 5(y + 4) = 0 \]

Now, we can factor out \((y + 4)\):

\[ (y + 4)(y + 5) = 0 \]

Setting each factor to zero gives us the possible values for \(y\):

1. \(y + 4 = 0 \implies y = -4\)
2. \(y + 5 = 0 \implies y = -5\)

Thus, the solutions for \(y\) are:

\[ y = -4 \quad \text{and} \quad y = -5 \]

### Summary of Solutions

The values we found are:
- For \(x\): \(x = -2\) and \(x = -\frac{1}{2}\)
- For \(y\): \(y = -4\) and \(y = -5\)