In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer.

I. $4x^2+20x+25=0$
II. $2y^2+11y+15=0$

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

The correct option is: If x ≥ y

Step 1: Solve Equation I for $x$

Equation I is given as:

$4x^2+20x+25=0$

Notice that this is a perfect square trinomial, as $4x^2={\left(2x\right)}^2$, $25=5^2$, and the middle term is $2\times \left(2x\right)\times 5=20x$.
Thus, we can rewrite the equation as:

${\left(2x+5\right)}^2=0$

Taking the square root on both sides:

$2x+5=0$

$2x=-5$

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

Step 2: Solve Equation II for $y$

Equation II is given as:

$2y^2+11y+15=0$

To solve this quadratic equation by splitting the middle term, we need two numbers whose product is $2\times 15=30$ and whose sum is $11$.
These numbers are $6$ and $5$.
Rewrite the middle term:

$2y^2+6y+5y+15=0$

Factor by grouping:

$2y\left(y+3\right)+5\left(y+3\right)=0$

$\left(2y+5\right)\left(y+3\right)=0$

Setting each factor to zero:

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

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

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

We have:
Values of $x$: $-2.5$
Values of $y$: $-2.5,-3$

Comparing each value:
1. When $x=-2.5$ and $y=-2.5$, $x=y$.
2. When $x=-2.5$ and $y=-3$, $x>y$ (since $-2.5$ is greater than $-3$).

Combining both comparisons, we get:

$x\ge y$