Hi there I am working thru this problem and my solution is (neg infinity, 3/2) Here is the problem : 2x^3-7x^2<-6x
My textbook has a different answer of x>3/2 am I wrong or is the book wrong?
I think that you are both mistaken.
If you factor you get
$\displaystyle 2x^3-7x^2+6x < 0 \iff x(x-2)\left(2x-3 \right)< 0$
This gives
$\displaystyle \begin{tabular}{|c|c|c|c|c|} \hline f(x) & - &+ &- &+ \\ \hline x & - & + & + & + \\ \hline (2x-3) & - & -& + & + \\ \hline (x-2) & - & - & - & + \\ \hline & (infinity,0) & (0,3/2) & ( 3/2, 2 ) & (2, infinity) \\ \hline \end{tabular}$
So the solution should be
$\displaystyle (-\infty,0) \cup \left( \frac{3}{2},2\right)$