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 is: x < y.

To solve the first quadratic equation for $x$:
$x^2+31x+84=0$
We need to find two numbers that multiply to 84 and add up to 31. These numbers are 28 and 3. Splitting the middle term, we get:
$x^2+28x+3x+84=0$
Now, group and factor the terms:
$x(x+28)+3(x+28)=0$
$(x+28)(x+3)=0$
Thus, the roots for the first equation are:
$x=-28\text{or}x=-3$

To solve the second quadratic equation for $y$:
$4y^2-19y+21=0$
We need to find two numbers that multiply to 84 (since 4 × 21 = 84) and add up to -19. These numbers are -12 and -7. Splitting the middle term, we get:
$4y^2-12y-7y+21=0$
Group and factor the terms:
$4y(y-3)-7(y-3)=0$
$(4y-7)(y-3)=0$
Thus, the roots for the second equation are:
$y=\frac{7}{4}=1.75\text{or}y=3$

Now we compare the values of $x$ and $y$:
- Comparing $x=-28$ with $y=1.75$ and $y=3$, we have $x<y$.
- Comparing $x=-3$ with $y=1.75$ and $y=3$, we also have $x<y$.
Since all values of $x$ are strictly less than all values of $y$, the final relationship is:
$x<y$