Found a real number $t$ such that $(14 + 5 \sqrt{3})(5 - \sqrt{3})\sqrt{8- 2 \sqrt{15}}= t \sqrt{2}$ In order to find the real number $t $ such that $A=(14 + 5 \sqrt{3})(5 - \sqrt{3})\sqrt{8- 2 \sqrt{15}}= t \sqrt{2}$ I have already compute this two expression \begin{align*} a=5+\sqrt{3} & = \sqrt{28 + 10\sqrt{3}} \\\ b=5-\sqrt{3} & = \sqrt{28 - 10\sqrt{3}} \end{align*} because $a^2 = 28 + 10\sqrt{3}$ and $b^2=28 - 10\sqrt{3}$. Then $2A=(28 + 10 \sqrt{3})(5 - \sqrt{3})\sqrt{8- 2 \sqrt{15}}= a^2b \sqrt{8- 2 \sqrt{15}}$. Additional calculus get me anywhere. Is there a charitable sool to give some help. Thank in advance for your comprehension.

Hint:

$$8-2\sqrt{15}=\cdots=(\sqrt5-\sqrt3)^2$$

We know for real $a,$ $$\sqrt{a^2}=|a|$$

which $=a,$ if $a\ge0$

else $=-a$

Now just multiply out