what will be the max revenue if I am given 800+60x-x^2

Assuming that what you've written is your Revenue (as a function of x units sold), first note that it's a quadratic with a negative leading coefficient. That means that its graph opens downward, and so its vertex represents its maximum value. Rewrite that quadratic into vertex form, and the y-coordinate of the vertex represents maximum Revenue. (And of course, the x-coordinate will tell you the number of units at which max Rev is achieved.)

Alternatively, take the first derivative of the Revenue function, and find the x-value which makes the derivative function equal to zero. Evaluate the Revenue function at this particular x-value, and you have your max Revenue.

(Or just for grins, do it both ways and make sure you produce the same result.)