quadratic equation expressed as a range of values 
Solve $6 - x - x^2 < 0$
A. $-3 < x < 2$
B. $x < -3, x > 2$
C. $-2 < x < 3$
D. $x < -2, x >3$

So this is a very easy quadratic to solve:
$x^2 + x - 6 > 0$
$(x + 3)(x -2) > 0$
What I struggle to in this question is to express the range.
Does anyone have any tips to expressing the range?
I would have said the answer is A.
 A: One thing to keep in mind, first of all, is the shape of the graph of
a second-degree polynomial in $x$.
Either the coefficient of $x^2$ is positive, in which case you get
a U-shaped parabola with a minimum value at the "bottom" of the U
and $y$-values that go toward $+\infty$ as you move $x$ away from
the minimum in either direction,
or the coefficeint of $x^2$ is negative
(as it is in this problem), in which case you get an
"upside-down" parabola with a maximum value
and $y$-values that go toward $-\infty$ as you move $x$ away from
the minimum in either direction.
So for the "upside-down" parabola, we always have negative values
of the polynomial for very negative $x$ or very positive $x$;
whether there are any positive values at all depends on whether there
are two roots, and if there are two roots, the positive values
are all between the roots.
Since you have $6 - x - x^2 < 0$, you are looking for negative values,
that is, values that are not between the roots.
You can prove that this is what you need by examining where the sign
changes of $(x+3)(x−2)$ occur and what the signs are between the changes,
but a good mental image of what the parabolic graph of the polynomial
looks like can help you avoid wrong answers.
A: $$f(x) = (x+3)(x-2)$$
We know that at $x = -3$ and $x = 2$ $f(x)$ is zero. Since the function is continuous, we can conclude that it changes sign only on these two points (since these are the points where it crosses the $x$-axis) . So just check what happens when $x < -3$, $-3 < x < 2$ and $2 < x$.

 When $x < -3$ we have that $x+3 <  0$ and $x -2 < 0$ and since product of two negatives is positive, we can say that $(x+3)(x-2) > 0, \text{for} \ x < -3$. Perform this analysis for the other ranges as well

Alternatively, we know that the function is monotonically increasing and that it crosses the $x$-axis at just two points. Can you use this argument to come up with an alternative approach to solve the problem?
A: I have this idea:
$$6-x-x^2=-(x^2+x-6)<0\iff(x+3)(x-2)>0\iff\begin{cases}x<-3\\{}\\\text{or}\\{}\\x>2\end{cases}$$
Just take into account that $\;-(x^2+x-6)\;$ is a downwards parabola with zeros at $\;x=-3,\,2\;$ and vertex at
$$\left(-\frac b{2a}\,,\,\,-\frac{\Delta}{4a}\right)=\left(-\frac12\,,\,\,\frac{25}4\right)$$
A: Notice that for all $x<-3$, 
$$(x+3)<0 \;\;\; \text{and} \;\;\; (x-2)<0$$
and it follows that
$$(x+3)(x-2)>0$$
Similarly for all $x>2$, 
$$(x+3)>0 \;\;\; \text{and} \;\;\; (x-2)>0$$
and it follows that
$$(x+3)(x-2)>0$$
Perform a similar analysis when $-3<x<2$.
A: The solution of the problem as your point out is that the function it is $0$ for
$x=-3$ and $x=2$. Then, you have to check if for $x<-3$, $x>2$ and $-3<x<2$, your results is negative of positive. If you check it you will get that:
$x\in (-\infty,3)\cup (2,\infty)$
