The function $f(x)=(1-\frac{\sqrt{21-4a-a^2}}{a+1})x^3+5x+\sqrt7$ is increasing at every point of its domain. Find the set of values of $a$ for which the function $f(x)=(1-\frac{\sqrt{21-4a-a^2}}{a+1})x^3+5x+\sqrt7$ is increasing at every point of its domain.

For $f(x)$ to be increasing,$f'(x)=3(1-\frac{\sqrt{21-4a-a^2}}{a+1})x^2+5\ge 0$
So,$(1-\frac{\sqrt{21-4a-a^2}}{a+1})\ge 0$ gives $(-\infty,-5)\bigcup(2,\infty)$ but the answer in my book is $[-7,-1)\bigcup[2,3]$

The domain gives $a^2+4a-21\leq0$ and $a\neq-1$, which gives $-7\leq a\leq3$ and $a\neq-1$.
Now, $1-\frac{\sqrt{21-4a-a^2}}{a+1}=0$ for $a=2$ and use the intervals method. 
Maybe the mistake was in  solving of the equation:
$$a+1=\sqrt{21-4a-a^2}.$$
We have $a>-1$ and $21-4a-a^2\geq0$, which gives $-1<a\leq3$.
Now, after squaring of the both sides we obtain $(a-2)(a+5)=0$, which gives $a=2$.
