Analyze both quadratic equations presented below to find the relationship between the variables, and select the correct option.

$\text{I.}2x^2-13x+20=0$

$\text{II.}{y}^2+10y+24=0$

The correct option is If x > y.

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

Step 1: Solve Equation I for $x$

The first quadratic equation is:

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

We split the middle term $-13x$ into two terms whose sum is $-13$ and whose product is $2\times 20=40$. The numbers are $-8$ and $-5$.

$2x^2-8x-5x+20=0$

$2x(x-4)-5(x-4)=0$

$(2x-5)(x-4)=0$

This yields two potential values for $x$:

$x=\frac{5}{2}=2.5$ or $x=4$

Step 2: Solve Equation II for $y$

The second quadratic equation is:

$y^2+10y+24=0$

We split the middle term $10y$ into two terms whose sum is $10$ and whose product is $24$. The numbers are $6$ and $4$.

$y^2+6y+4y+24=0$

$y(y+6)+4(y+6)=0$

$(y+6)(y+4)=0$

This yields two potential values for $y$:

$y=-6$ or $y=-4$

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

The values of $x$ are $2.5$ and $4$, both of which are positive.

The values of $y$ are $-6$ and $-4$, both of which are negative.

Comparing each value of $x$ with each value of $y$:

1. When $x=2.5$, $2.5>-6$ and $2.5>-4$ ($x>y$)

2. When $x=4$, $4>-6$ and $4>-4$ ($x>y$)

In all cases, $x>y$.