In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer
I `3x^2+14x+15=0`
II `2y^2+9y+10=0`

Options

A. if`xgey`
B. if`xlty`
C. if`xley`
D. if x =y or the relationship between x and y cannot be establised

Answer

To solve the given equations step by step, we will first solve each equation separately and then compare the values of \(x\) and \(y\).

### Step 1: Solve Equation I
The first equation is:
\[ 3x^2 + 14x + 15 = 0 \]

To solve this quadratic equation, we can use the factorization method. We need to find two numbers that multiply to \(3 \times 15 = 45\) and add to \(14\).

The factors of \(45\) that add up to \(14\) are \(9\) and \(5\). Therefore, we can rewrite the middle term:
\[ 3x^2 + 9x + 5x + 15 = 0 \]

Now, we can group the terms:
\[ (3x^2 + 9x) + (5x + 15) = 0 \]

Factoring by grouping:
\[ 3x(x + 3) + 5(x + 3) = 0 \]

Now factor out the common term \((x + 3)\):
\[ (3x + 5)(x + 3) = 0 \]

Setting each factor to zero gives us:
1. \(3x + 5 = 0 \Rightarrow x = -\frac{5}{3} \approx -1.67\)
2. \(x + 3 = 0 \Rightarrow x = -3\)

### Step 2: Solve Equation II
The second equation is:
\[ 2y^2 + 9y + 10 = 0 \]

Again, we will factor this quadratic equation. We need two numbers that multiply to \(2 \times 10 = 20\) and add to \(9\).

The factors of \(20\) that add up to \(9\) are \(5\) and \(4\). We can rewrite the middle term:
\[ 2y^2 + 5y + 4y + 10 = 0 \]

Now, we can group the terms:
\[ (2y^2 + 5y) + (4y + 10) = 0 \]

Factoring by grouping:
\[ y(2y + 5) + 2(2y + 5) = 0 \]

Now factor out the common term \((2y + 5)\):
\[ (2y + 5)(y + 2) = 0 \]

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

### Step 3: Compare Values of \(x\) and \(y\)
Now we have the possible values for \(x\) and \(y\):
- For \(x\): \(-\frac{5}{3} \approx -1.67\) and \(-3\)
- For \(y\): \(-\frac{5}{2} = -2.5\) and \(-2\)

Now we will compare:
1. Comparing \(x = -\frac{5}{3} \approx -1.67\) with \(y = -2\):
\[ -1.67 > -2 \quad \text{(So, } x > y\text{)} \]

2. Comparing \(x = -3\) with \(y = -2.5\):
\[ -3 < -2.5 \quad \text{(So, } x < y\text{)} \]

### Conclusion
Since we have found that \(x\) can be greater than \(y\) in one case and less than \(y\) in another case, we cannot establish a consistent relationship between \(x\) and \(y\).

Thus, the answer is:
**The relationship between \(x\) and \(y\) cannot be established.**