$\text{(I)}{x}^2+5x+6=0$

$\text{(II)}{y}^2-5y+6=0$

The correct answer is If x < y.

Let's solve both quadratic equations step-by-step to find the values of $x$ and $y$ and determine their relationship.

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

$x^2+5x+6=0$

To factor this quadratic equation, we look for two numbers that multiply to 6 and add up to 5. These numbers are 2 and 3.
$x^2+2x+3x+6=0$
$x(x+2)+3(x+2)=0$
$(x+2)(x+3)=0$

Setting each factor equal to zero gives:
$x=-2$ or $x=-3$

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

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

To factor this equation, we need two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3.
$y^2-2y-3y+6=0$
$y(y-2)-3(y-2)=0$
$(y-2)(y-3)=0$

Setting each factor equal to zero gives:
$y=2$ or $y=3$

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

The values of $x$ are $-2$ and $-3$ (both negative).
The values of $y$ are $2$ and $3$ (both positive).

Since any negative number is strictly less than any positive number:
$-2<2$
$-2<3$
$-3<2$
$-3<3$

Thus, in all cases, $x<y$.