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 $x$ and $y$, we need to solve both quadratic equations step-by-step.

Step 1: Solve Equation I for $x$

The first quadratic equation is:

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

To factorize this quadratic expression, we look for two numbers that multiply to give $2\times 20=40$ and add up to give $-13$. These two numbers are $-5$ and $-8$.

Splitting the middle term:

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

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

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

Setting each factor to zero:

$2x-5=0\Rightarrow x=\frac{5}{2}=2.5$

$x-4=0\Rightarrow x=4$

Thus, the values of $x$ are $2.5$ and $4$.

Step 2: Solve Equation II for $y$

The second quadratic equation is:

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

To factorize this quadratic expression, we look for two numbers that multiply to give $24$ and add up to give $10$. These two numbers are $6$ and $4$.

Splitting the middle term:

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

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

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

Setting each factor to zero:

$y+6=0\Rightarrow y=-6$

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

Thus, the values of $y$ are $-6$ and $-4$.

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

We compare each value of $x$ with every value of $y$:

1. Comparing $x=2.5$ with $y=-6$: $2.5>-6$ ($x>y$)

2. Comparing $x=2.5$ with $y=-4$: $2.5>-4$ ($x>y$)

3. Comparing $x=4$ with $y=-6$: $4>-6$ ($x>y$)

4. Comparing $x=4$ with $y=-4$: $4>-4$ ($x>y$)

In all cases, both values of $x$ are strictly greater than both values of $y$. Therefore, the correct relationship between the variables is $x>y$.