$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})$ 
Let $k$ be equal to:
  $$k=-\sqrt{3} \left(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}} \right)$$

I am trying to simplify the expression and express $5+2\sqrt{6}$ and $5-2\sqrt{6}$ as squares. I don't think it's smart to multiply the brackets by $-\sqrt{3}$ at first. Is there any algorithm that I should know to express the expressions as squares?

You have $\sqrt{A \pm \sqrt{B}}=\sqrt{\frac{A+C}{2}}\pm \sqrt{\frac{A-C}{2}}$ where $C=\sqrt{A^2-B}$
So we can write $\sqrt{5+2\sqrt{6}}=\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+1}{2}}+\sqrt{\frac{5-1}{2}}=\sqrt{3}+\sqrt{2}$
and $\sqrt{5-2\sqrt{6}}=\sqrt{5-\sqrt{24}}=\sqrt{\frac{5+1}{2}}-\sqrt{\frac{5-1}{2}}=\sqrt{3}-\sqrt{2}$
So substituting we get $k=-\sqrt{3}(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2})=-6$
