$a+b\sqrt{3}=\sqrt{21-12\sqrt{3}}, a,b \in \mathbb {Z}$ Find a+b So far I've reasoned that $\mathbf{a}$ and $\mathbf{b}$ can't be both negative, because $\sqrt{21-12\sqrt{3}}$ cannot be negative. 
Also $\mathbf{a}$ and $\mathbf{b}$ can't be both positive, because $\sqrt{21-12\sqrt{3}}$  is from 0 to 1, thus there is no positive whole numbers which could satisfy that $\mathbf{a}$ plus $\mathbf{b}*\sqrt{3}$ is close to 0 and 1.
At this stage, I don't know what to do. I appreciate any help.

Write 
$$\left(a+b\sqrt{3}\right)^2=a^2+3b^2+2ab\sqrt{3}=21-12\sqrt{3}$$
This means
$$\begin{cases}
a^2+3b^2&=21\\
2ab&=-12
\end{cases}$$
Now substitute $a=-6/b$. You get the following biquadratic
$$b^4-7b^2+12=0$$
Has two solution for $b^2$, $3$ to be discarded because $b$ is an integer and $4$ which means that
$$b=\pm 2\\a=\pm 3$$
The solution to retain is the non negative one ie $-3+2\sqrt{3}$
