Simplifying $\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7}$ 
$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$

I'm having trouble solving this equation. I've tried squaring both sides and got this
$$11 - 4\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} + 8 - 2\sqrt{7} = a^2$$
after simplifying
$$19 - 6\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} = a^2$$
and that's where I got stuck.

There is a formula to simplify nested radicals like that
$$\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
It is effective only if $(a^2-b)$ is a perfect square. In your example both of them have this property
$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$
$$\sqrt {11 -  \sqrt {112}} - \sqrt {8 -\sqrt {28}} = a$$
$$\sqrt {11 -  \sqrt {112}} =\sqrt{\frac{11+\sqrt{121-112}}{2}}-\sqrt{\frac{11-\sqrt{121-112}}{2}}=\sqrt{\frac{11+3}{2}}-\sqrt{\frac{11-3}{2}}=$$
$$=\sqrt 7-2$$
$$\sqrt {8 -\sqrt {28}} =\sqrt{\frac{8+6}{2}}-\sqrt{\frac{8-6}{2}}=\sqrt 7-1$$
so finally
$$\sqrt 7-2-(\sqrt 7-1)=-1$$

There is a similar formula for the sum
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
