Question 763316
By cheating (not saying how), this seems to have zeros at {{{x=1&1/3}}} and {{{x=-1&1/2}}}.  Try Rational Roots theorem and synthetic division to see how these roots may be.  


Also, since your expression polynomial is degree 2, why not directly use general quadratic formula solution?  One way to look at this is that multiplying by zero does not change anything to nonzero.  You have these:


{{{x^2+(1/6)x-2=0}}}
multiplying both sides by 6 you get...


{{{6x^2+1x-12=0}}}, which is EQUIVALENT.

Both of those have the same zeros or same solution.