You do not need a caluclator for this problem. Group the first two terms together and the second two terms togerter. Factor out x2 from the first two terms and -5 from the second two terms:
(x3+4x2)+(-5x-20) = x2(x+4)-5(x+4)
Because x+4 is a factor of both groups, it can also be factored out:
x2(x+4)-5(x+4) = (x2-5)(x+4)
Setting g(x)=0, the three values of x which solve this equation are -4, and ±√5.
The way you solve this equation on a calculator can be done many ways. Using the Ti-89, one could simply use the solve function (F2:1). This arugement must have an expression, a defined variable.
I beleive the zeros can be solved for graphically as well on other graphing calculators. Enter the equation using the 'Y=' menu; fix the window so that the function is visable on the screen [-5,5]x[-20,20]; open the math menu and select the zero function. You will have to define a lower bound and an upper bound (a point before the desired zero and a point after the zero). Keep in mind that the zero function solves for one zero at a time.