Answer
See below.
Work Step by Step
Let's compare $f(x)=-3x^2+120x+50$ to $f(x)=ax^2+bx+c$. We can see that a=-3, b=120, c=50. $a\lt0$, hence the graph opens down, hence its vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{120}{2\cdot(-3)}=20.$ Hence the maximum value is $f(20)=-3(20)^2+120(20)+50=-1200+2400+50=1250.$
