The number of real roots of $\frac{2x-3}{x-1}+1=\frac{6x^2-x-6}{x-1}$ is

$3x-4=6x^2-x-6$

$3x^2-2x-1=0$

$\therefore x=1\left(or\right)x=-\frac{1}{3}$

But at $x=1$ given equation doesnot exist

$\therefore x=\frac{-1}{3}$