Directions:In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer.
I. `3x^2-4x-4=0` II. `4y^2+23y+15=0`

Options

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

Answer

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

### Step 1: Solve the first equation \(3x^2 - 4x - 4 = 0\)

We will use the quadratic formula to solve for \(x\):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 3\), \(b = -4\), and \(c = -4\).

#### Step 1.1: Calculate the discriminant
\[ b^2 - 4ac = (-4)^2 - 4 \cdot 3 \cdot (-4) = 16 + 48 = 64 \]

#### Step 1.2: Substitute into the quadratic formula
\[ x = \frac{-(-4) \pm \sqrt{64}}{2 \cdot 3} = \frac{4 \pm 8}{6} \]

#### Step 1.3: Calculate the two possible values for \(x\)
1. \(x_1 = \frac{4 + 8}{6} = \frac{12}{6} = 2\)
2. \(x_2 = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3}\)

### Step 2: Solve the second equation \(4y^2 + 23y + 15 = 0\)

Again, we will use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 4\), \(b = 23\), and \(c = 15\).

#### Step 2.1: Calculate the discriminant
\[ b^2 - 4ac = (23)^2 - 4 \cdot 4 \cdot 15 = 529 - 240 = 289 \]

#### Step 2.2: Substitute into the quadratic formula
\[ y = \frac{-23 \pm \sqrt{289}}{2 \cdot 4} = \frac{-23 \pm 17}{8} \]

#### Step 2.3: Calculate the two possible values for \(y\)
1. \(y_1 = \frac{-23 + 17}{8} = \frac{-6}{8} = -\frac{3}{4}\)
2. \(y_2 = \frac{-23 - 17}{8} = \frac{-40}{8} = -5\)

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

We have the solutions:
- For \(x\): \(2\) and \(-\frac{2}{3}\)
- For \(y\): \(-\frac{3}{4}\) and \(-5\)

Now we compare:
1. \(2\) is greater than both \(-\frac{3}{4}\) and \(-5\).
2. \(-\frac{2}{3}\) is greater than \(-5\) but less than \(-\frac{3}{4}\).

### Conclusion
Since \(x = 2\) is greater than both values of \(y\), we conclude that \(x\) is greater than \(y\).

### Final Answer
The relation is \(x > y\).

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