How many integers $x$ satisfy: $-4How many integers are there in the set $\{x\in\Bbb R\mid-4<x^2+5x<14\}$?
The correct answer is four, but I only find $0$ and $1$ as integers in the set. 
Is the answer wrong?
 A: Let's repose the constraint:
$$
-4<x^2+5x<14\iff-16<4x^2+20x<56\iff 9<4x^2+20x+25<81.
$$
Recognizing squares, we have
$$
3^2<(2x+5)^2<9^2\iff 2x+5=\pm 4,\pm 5,\pm 6,\pm 7,\pm 8.
$$
Note you want integer $x$ so $2x+5$ is odd, leaving 4 possible values for $2x+5$: $\pm 5, \pm 7$, corresponding to $4$ values of $x$: $-6,-5,0,1$. 
A: This looks like a homework problem with the following intended solution:
Consider $-4<x^2+5x<14$ one inequality at a time.
(In each case, it may be helpful to draw a graph of the corresponding parabola.)
First, $x^2 + 5x > -4$ iff $(x+1)(x+4) = x^2 + 5x + 4 > 0$.
So, we have the points $x < -4$ or $x > -1$.
Second, $x^2 + 5x < 14$ iff $(x+7)(x-2) = x^2 + 5x - 14 < 0$.
So, we have the points $-7<x<2$. 
Among integers, this means considering $\{-6, -5, -4, -3, -2, -1, 0, 1\}$.
Within this set, and returning to the first inequality, which integers are less than $-4$?
Answer: $-6$ and $-5$.
Similarly, which integers are greater than $-1$?
Answer: $0$ and $1$.
Thus, the answer to your question is the four integers $\{-6, -5, 0, 1\}$. 
