In the following question, two quadratic equations labeled I and II are provided. Solve both equations to determine the relationship between x and y.

I. x² - 7x - 260 = 0
II. y² - 3y - 54 = 0

The correct option is If x = y or no relation can be established.

Step 1: Solve Equation I for x.

$x^2-7x-260=0$

To factorize the equation, find two numbers whose product is -260 and sum is -7. These numbers are -20 and +13.

$x^2-20x+13x-260=0$

$x(x-20)+13(x-20)=0$

$(x-20)(x+13)=0$

Hence, the values of x are:

$x=20$ or $x=-13$

Step 2: Solve Equation II for y.

$y^2-3y-54=0$

To factorize the equation, find two numbers whose product is -54 and sum is -3. These numbers are -9 and +6.

$y^2-9y+6y-54=0$

$y(y-9)+6(y-9)=0$

$(y-9)(y+6)=0$

Hence, the values of y are:

$y=9$ or $y=-6$

Step 3: Compare the values of x and y.
- For $x=20$ and $y=9$: $x>y$
- For $x=20$ and $y=-6$: $x>y$
- For $x=-13$ and $y=9$: $x<y$
- For $x=-13$ and $y=-6$: $x<y$

Since in some cases $x>y$ and in other cases $x<y$, no relation can be established between x and y.