AlgebraII factoring polynomials equation: $2x^2 - 11x - 6$
Using the quadratic formula, I have found the zeros: $x_1 = 6, x_2 = -\frac{1}{2}$
Plug the zeros in: $2x^2 + \frac{1}{2}x - 6x - 6$
This is where I get lost. I factor $-6x - 6$ to: $-6(x + 1)$, but the answer says otherwise. I am also having trouble factoring the left side.
Could someone please explain to me why the answer to the question was: $(x - 6)(2x + 1)$. How does $-\frac{1}{2}$ become $1$?

$$
2x^2-11x - 6 = 2(x -\bullet)(x-\bullet)
$$
The two "$\bullet$"s are the two roots, $6$ and $\dfrac{-1}2$.  So you have
$$
2(x-6)\left(x-\frac {-1} 2 \right) = (x-6)\  \underbrace{2\left(x+\frac 1 2 \right)} = (x-6)(2x+1).
$$
So $2\left(x + \dfrac 1 2\right)$ becomes $2x+1$.
