solve the equation $\displaystyle 6x^3-19x^2+11x+6=0$ so far i got (x-2) as a factor thats as far as i can seem to get
Actually, when you solve by factoring $\displaystyle 6x^2-7x-3=0$, you get:
$\displaystyle (3x+1)(2x-3)=0$
So, the roots are:
$\displaystyle x=-\frac{1}{3}$ and $\displaystyle x=\frac{3}{2}$.
So, one of your roots was correct. You don't want to round off the other one to -.3
Leave it as a fraction.