Two quadratic equations, labeled (I) and (II), are given below. Solve both equations to determine the relationship between x and y:

$\text{I.}{x}^2-11x+28=0$

$\text{II.}{y}^2+13y+30=0$

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

Step 1: Solve Equation (I) for $x$

The first equation is:

$x^2-11x+28=0$

We look for two numbers that multiply to give $28$ and add up to $-11$. These numbers are $-7$ and $-4$.

Factoring the equation gives:

$(x-7)(x-4)=0$

Setting each factor to zero:

$x-7=0\Rightarrow x=7$

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

Thus, the values of $x$ are $4$ and $7$.

Step 2: Solve Equation (II) for $y$

The second equation is:

$y^2+13y+30=0$

We look for two numbers that multiply to give $30$ and add up to $13$. These numbers are $10$ and $3$.

Factoring the equation gives:

$(y+10)(y+3)=0$

Setting each factor to zero:

$y+10=0\Rightarrow y=-10$

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

Thus, the values of $y$ are $-10$ and $-3$.

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

Let's compare every value of $x$ with every value of $y$:
- Comparing $x=4$ with $y=-10$: $4>-10$
- Comparing $x=4$ with $y=-3$: $4>-3$
- Comparing $x=7$ with $y=-10$: $7>-10$
- Comparing $x=7$ with $y=-3$: $7>-3$

In all cases, every possible value of $x$ is strictly greater than every possible value of $y$ ($x>y$).

Therefore, the correct answer is If x is greater than y.