In the following question, two equations numbered I and II are given. Solve both equations and select the correct option expressing the relationship between x and y.

$\begin{array}{ll}\text{I.}& 2x^2-9x+10=0\\ \text{II.}& y^2+7y+12=0\end{array}$

The correct answer is If x > y.

To determine the relationship between $x$ and $y$, we solve both quadratic equations step-by-step.

Step 1: Solve Equation I for $x$

The first equation is:

$2x^2-9x+10=0$

Splitting the middle term:

$2x^2-4x-5x+10=0$

Factoring by grouping:

$2x(x-2)-5(x-2)=0$

$(2x-5)(x-2)=0$

Solving for $x$:

$2x-5=0\Rightarrow x=2.5$

$x-2=0\Rightarrow x=2$

Thus, the roots for $x$ are $2$ and $2.5$.

Step 2: Solve Equation II for $y$

The second equation is:

$y^2+7y+12=0$

Splitting the middle term:

$y^2+3y+4y+12=0$

Factoring by grouping:

$y(y+3)+4(y+3)=0$

$(y+3)(y+4)=0$

Solving for $y$:

$y+3=0\Rightarrow y=-3$

$y+4=0\Rightarrow y=-4$

Thus, the roots for $y$ are $-3$ and $-4$.

Step 3: Compare the values of $x$ and $y$

Comparing the roots:

For $x=2$:
$2>-3$ and $2>-4$

For $x=2.5$:
$2.5>-3$ and $2.5>-4$

Since both values of $x$ are strictly greater than both values of $y$, we have $x>y$.