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

$\text{I. }{x}^2+31x+84=0$

$\text{II. }4y^2-19y+21=0$

The correct answer/option is x < y.

To compare the values of $x$ and $y$, we need to solve both quadratic equations separately.

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

The first equation is:
$x^2+31x+84=0$
To factorize this quadratic equation, we find two numbers whose product is $84$ and whose sum is $31$. These numbers are $28$ and $3$ (since $28\times 3=84$ and $28+3=31$).

We can rewrite the equation as:
$x^2+28x+3x+84=0$
$x(x+28)+3(x+28)=0$
$(x+28)(x+3)=0$
Solving for $x$ gives:
$x=-28$ or $x=-3$

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

The second equation is:
$4y^2-19y+21=0$
To factorize this quadratic equation, we find two numbers whose product is $4\times 21=84$ and whose sum is $-19$. These numbers are $-12$ and $-7$ (since $-12\times -7=84$ and $-12+(-7)=-19$).

We can rewrite the equation as:
$4y^2-12y-7y+21=0$
$4y(y-3)-7(y-3)=0$
$(4y-7)(y-3)=0$
Solving for $y$ gives:
$y=\frac{7}{4}=1.75$ or $y=3$

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

Comparing the solution sets:
$x\in \{-28,-3\}$
$y\in \{1.75,3\}$
Since both possible values of $x$ ($-28$ and $-3$) are strictly less than both possible values of $y$ ($1.75$ and $3$), we conclude that:
$x<y$