If you mean g(f(x))=16 then:
g(f(x))=9-(2x-3)2=9-4x2+12x-9=16, or we can write:
g(f(x))=(3-2x+3)(3+2x-3)=2x(6-2x)=12x-4x2=16. Either way:
4x2-12x+16=0=4(x2-3x+4). This has complex roots only:
x=(3±√(9-16))/2=(3±i√7)/2. So there are no real roots.
If you mean g×f(x)=(9-x2)(2x-3) then -2x3+3x2+18x-27=16,
2x3-3x2-18x+43=0, which has 2 complex roots and one real root. So this can be solved for a real solution using various methods.
In each case we get an equation and there are solutions, but not necessarily real solutions.