jmc

We raise both sides to the fourth power, which is equivalent to squaring twice, in order to get rid of the radicals. The left-hand side becomes $${\left(\sqrt{2\sqrt{t-2}}\right)}^4={\left(2\sqrt{t-2}\right)}^2=4\cdot (t-2)=4t-8.$$The right-hand side becomes ${\left(\sqrt[4]{7-t}\right)}^4=7-t$. Setting them equal, $$4t-8=7-t⟹5t=15,$$and $t=\boxed{3}$. Checking, we find that this value does indeed satisfy the original equation.