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 $1$?

When you factor out the equation $2x^2-11x-6$, you get $(x-6)(2x+1)$ (David Peterson did the factoring process). This shows that the functions has two zeros in the graph. Thus, we have to sent $(x-6)(2x+1)$ equal to $0$: $$(x-6)(2x+1)=0.$$ Then we find the zeros: 
$$(x-6)(2x+1)=0$$ 

$$x-6=0$$ 
$$\boxed{x=6}$$ 

$$2x+1=0$$ 
$$2x=-1$$ 
$$\boxed{x=-\frac{1}{2}}.$$ 
Therefore, we can clearly see that by factoring and sovling for $x$, we get  $$\boxed{x=6,-\frac{1}{2}}.$$
