jmc

Let $a=\sqrt[4]{47-2x}$ and $b=\sqrt[4]{35+2x}.$ Then $a+b=4.$ Also, $$a^4+b^4=(47-2x)+(35+2x)=82.$$Since $a+b=4,$ there exists a $t$ such that $a=2+t$ and $b=2-t.$ Then $$a^4+b^4=(2+t{)}^4+(2-t{)}^4=2t^4+48t^2+32=82.$$This simplifies to $t^4+24t^2-25=0$, which factors as $(t^2-1)(t^2+25)=0.$ Hence, $t=\pm 1.$

If $t=1,$ then $a=\sqrt[4]{47-2x}=3,$ which leads to $x=-17.$ If $t=-1,$ then $a=\sqrt[4]{47-2x}=1,$ which leads to $x=23.$ Thus, the solutions are $\boxed{23,-17}.$ We check that these solutions works.