Answer: \(\frac{2}{5} \lt x \lt \frac{3}{2}\).

Expand the right hand side first:

\((2x+1)(x-1)=2x^2-x-1.\)

So the inequality becomes

\(12x^2-20x+5 \lt 2x^2-x-1.\)

Bring all terms to the left:

\(10x^2-19x+6 \lt 0.\)

Factorise:

\(10x^2-19x+6=(5x-2)(2x-3).\)

Hence

\((5x-2)(2x-3)\lt 0.\)

The critical values are \(x=\frac25\) and \(x=\frac32\). Since the quadratic has a positive coefficient of \(x^2\), it is negative between the roots.

Therefore

\(\boxed{\frac25\lt x\lt \frac32}.\)