consider:

assume(k[f1]>0,k[f2]>0,k[f2]>k[f1],h_bar>0,m>0);

h_bar:=1.0545e-34;m:=0.10938e-31;n[0]=1e28;

> eq1:=n=(k[f1]^3+k[f2]^3)/6/Pi^2;

> eq2:=e*V=h_bar^2/2/m*(k[f2]^2-k[f1]^2);

> solve({eq1,eq2},{k[f1],k[f2]});

in the final command i get a very messy numerical&symbolic results like

{k[f1] = 1016612041.*
(-1.*RootOf(9456017282782496601177464289*n^2*Pi^4-3152005760927498867059154763*n*Pi^2*_Z^3
+2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000*e^3*V^3
-3049836123000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000*e^2*V^2*_Z^2
+1550250062859278521500000000000000000000000000000000000000000000*e*V*_Z^4)^3
+59.21762641*n)/
(1016612041.*RootOf(9456017282782496601177464289*n^2*Pi^4
-3152005760927498867059154763*n*Pi^2*_Z^3
+2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000*e^3*V^3
-3049836123000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000*e^2*V^2*_Z^2
+1550250062859278521500000000000000000000000000000000000000000000*e*V*_Z^4)^2
-.2000000000e46*e*V),
k[f2] = RootOf(9456017282782496601177464289*n^2*Pi^4-3152005760927498867059154763*n*Pi^2*_Z^3
+2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000*e^3*V^3
-3049836123000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000*e^2*V^2*_Z^2
+1550250062859278521500000000000000000000000000000000000000000000*e*V*_Z^4)}

Is there anyway to make the result displayed in scientific notation like 2*10^50, instead of displaying all the 0s?