Factor the expression by grouping. First, the expression needs to be rewritten as $y^{2}+ay+by+12$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-7$$ $$ab=1\times 12=12$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $12$.
$$-1,-12$$ $$-2,-6$$ $$-3,-4$$
Calculate the sum for each pair.
$$-1-12=-13$$ $$-2-6=-8$$ $$-3-4=-7$$
The solution is the pair that gives sum $-7$.
$$a=-4$$ $$b=-3$$
Rewrite $y^{2}-7y+12$ as $\left(y^{2}-4y\right)+\left(-3y+12\right)$.
$$\left(y^{2}-4y\right)+\left(-3y+12\right)$$
Factor out $y$ in the first and $-3$ in the second group.
$$y\left(y-4\right)-3\left(y-4\right)$$
Factor out common term $y-4$ by using distributive property.
