Question 116758
This is not an equation.  This is a polynomial expression.  As it stands, there is no solution -- you can't 'solve' a polynomial expression. 


So, let's presume that you actually meant to say 'Please solve: {{{9x^2-17x^4+21x^5=0}}}.


The first thing you can do is factor {{{x^2}}} out of the left side leaving you with:


{{{x^2(21x^3-17x^2+9)=0}}}


That gives us two of the five roots directly, namely {{{x=0}}} and {{{x=0}}}


From here on it gets mighty ugly.   Looking at a graph of the cubic factor, we can see that we have one real root, slightly smaller than {{{-(1/2)}}} and a conjugate pair of complex roots.


{{{graph(600,600,-4,4,-20,20,21x^3-17x^2+9)}}}


My recommendation, to the extent that your instructor will be satisfied with a reasonable approximation, is to go to a website that has a general cubic equation solver.  When you use such a thing, just be sure to remember that the c coefficient in {{{ax^3+bx^2+cx+d=0}}} has a zero value in this problem.


There is another possibility that occurred to me -- you may have a typo in your polynomial.  If you actually meant {{{9x^3-17x^4+21x^5}}}, then {{{9x^3-17x^4+21x^5=0}}} becomes:


{{{x^3(21x^2-17x+9)}}} and we are left with a much easier to handle quadratic.


 *[invoke quadratic "x", 21, -17, 9]


Not nearly as messy, even though the roots are complex.



Hope this helps.