First, find #f(2)# by substituting every occurrence of #color(red)(n)# with #color(red)(2)# in the function #f(n)# and calculate the result:
#f(color(red)(n)) = 2color(red)(n)^2 + 5# becomes:
#f(color(red)(2)) = (2 * color(red)(2)^2) + 5#
#f(color(red)(2)) = (2 * 4) + 5#
#f(color(red)(2)) = 8 + 5#
#f(color(red)(2)) = 13#
Now, we know #g(f(2)) = g(13)#
To find #g(13)# substitute every occurrence of #color(red)(n)# with #color(red)(13)# in the function #g(n)# and calculate the result:
#g(color(red)(n)) = 3color(red)(n) + 2# becomes:
#g(color(red)(13)) = (3 * color(red)(13)) + 2#
#g(color(red)(13)) = 39 + 2#
#g(color(red)(13)) = 41#
#g(f(2)) = 41#