a) (f○g)(x) = f(g(x)) = 3g(x)+3 = 3(x²+4x)+3 = 3x²+12x+3
=> (f○g)(4) = 3·4²+12·4+3 = 99
b) (f○g)(x) = 0 => 3x²+12x+3 = 0
3x²+12x+3 = 0 |:3
x²+4x+1 = 0
delta = 4²-4·1·1 = 16-4 = 12
x1 = (-4+√12)/2 = (-4+2√3)/2 = -2+√3
x2 = (-4-√12)/2 = (-4-2√3)/2 = -2-√3
