To find #g(h(t))# we must substitute #color(red)(h(t))# or #color(red)(2t)# for each occurrence of #color(red)(t)# in #g(t) = -3t^3 - 5t^2#
#g(color(red)(t)) = -3color(red)(t)^3 - 5color(red)(t)^2# becomes:
#g(color(red)(h(t))) = -3color(red)((2t))^3 - 5color(red)((2t))^2#
#g(color(red)(h(t))) = (-3 * color(red)(2)^3 * color(red)(t)^3) - (5 * color(red)(2)^2 * color(red)(t)^2)#
#g(color(red)(h(t))) = (-3 * 8 * t^3) - (5 * 4 * t^2)#
#g(color(red)(h(t))) = 34t^3 - 20t^2#