Factor the expression by grouping. First, the expression needs to be rewritten as $y^{2}+ay+by-120$. To find $a$ and $b$, set up a system to be solved.
$$a+b=2$$ $$ab=1\left(-120\right)=-120$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-120$.
Rewrite $y^{2}+2y-120$ as $\left(y^{2}-10y\right)+\left(12y-120\right)$.
$$\left(y^{2}-10y\right)+\left(12y-120\right)$$
Factor out $y$ in the first and $12$ in the second group.
$$y\left(y-10\right)+12\left(y-10\right)$$
Factor out common term $y-10$ by using distributive property.
$$\left(y-10\right)\left(y+12\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$y^{2}+2y-120=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
