Question 138394
Your question is more correctly put: "What are the answers to..."  Since you have a cubic, or 3rd degree polynomial equation, the Fundamental Theorem of Algebra tells us there must be three roots.


In general terms, a cubic equation is a rather tiresome thing to have to solve.  Fortunately, however, your equation has a pair of common factors in every term:


{{{2y^3-18y^2+28y=0}}}


You can factor out a {{{2}}} and a {{{y}}} from every term:


{{{2y(y^2-9y+14)=0}}}


You can divide both sides by 2 to get:


{{{y(y^2-9y+14)=0}}}


Now, the Zero Product Rule tells us that either {{{y=0}}} or {{{y^2-9y+14=0}}}


But notice that {{{y^2-9y+14=0}}} is just a quadratic equation, and an easily factorable one at that. 


So, 0 is one of the elements of your solution set, and the roots of the quadratic are the other two.