Number of integral values of ‘a’ for which the quadratic equation $(\mathrm{a}-1){\mathrm{x}}^2+(4\mathrm{a}+1)\mathrm{x}-6=0$ has both roots integers is

$$\begin{gathered} {\mathrm{x}}^2+\left(\frac{4\mathrm{a}+1}{\mathrm{a}-1}\right)\mathrm{x}-\frac{6}{\mathrm{a}-1}=0 \\ \frac{4\mathrm{a}+1}{\mathrm{a}-1}=4+\frac{5}{\mathrm{a}-1}\in \mathrm{I}\to \mathrm{a}=0,2,6,-4 \\ \mathrm{Now}, -\frac{6}{\mathrm{a}-1}\in \mathrm{I}\mathrm{if}\mathrm{a}=0,2 \\ \mathrm{D}={\left(\frac{4\mathrm{a}+1}{\mathrm{a}-1}\right)}^2+\frac{24}{\mathrm{a}-1} \\ \mathrm{If} \mathrm{a}=0,\mathrm{\Delta }=1-24=-23<0 \\ \mathrm{If} \mathrm{a}=2,\mathrm{\Delta }=81+24=105 \mathrm{Not} \mathrm{perfect} \mathrm{square} \\ \therefore \mathrm{No} \mathrm{such} \mathrm{value} \mathrm{of} ‘\mathrm{a}’ \end{gathered}$$