Simplifying $\{ x \in \mathbb{R} : x^2 - x - 6 \geq 0 \}$ when possible Simplify the following interval notation when possible:
$$\{ x \in \mathbb{R} : x^2 - x - 6 \geq 0 \}$$
 A: Hint: 
Find the roots (there are 2) and than think about the behaviour of $x^2-x-6$ as $|x|$ tends to infty.
A: Here is the graph, look for the points where parabola is non-negative (y ≥ 0).
A: Read about Quadratic equation. That should give you a good idea of them, then you'll be able to find the roots. According to the roots you've found, you can figure out where the equation is positive or negative. 
Generally, $ax^2+bx+c$ has the opposite sign of $a$ between the roots. 
The given equation has $3$ and $-2$ as roots, then where it's greater than or equal to zero?
A: This is a bit of an extension to Americo Tavares' answer. 
First find any root, $x=3$ is fine. Now what you need is the second polynomial:
$$
\frac{x^2 -x-6}{x-3}=Q(x)
$$
where $Q(x)$ is of the form $ax+b$, the RHS becomes 
$$
Q(x)(x-3)=ax^2+(b-3a)x-3b
$$
Then you equate coefficients and fin that $a=1,b=2$, so the original polynomial can be factored as 
$$
P(x)=(x-3)(x+2) \geq 0
$$
Clearly the product is larger than 0 either when bothterms are positive or when both terms are negative, hence your solution is $x \leq -2 \cup x \geq 3$
