exponential equation $$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$
So I have squared both sides and got:
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$
I don't know what to do now

You've already seen that $(5-2\sqrt{6})(5+2\sqrt{6})=1$ when you squared both sides.  This means that $5-2\sqrt{6}=\frac{1}{5+2\sqrt{6}}$, so your last equation can be rewritten as $$\left(\frac{1}{5+2\sqrt6}\right)^x+(5+2\sqrt6)^x+2=100$$
or, letting $y=(5+2\sqrt{6})^x$,
$$\frac{1}{y}+y+2=100$$
so
$$1+y^2+2y=100y$$
which is quadratic in $y$.  Solve this for $y$, then use that solution to solve for $x$.
