$\text{Number of integral values of x satisfyng the inequality }{\left( \frac{3}{4}\right)}^{6x+10-x^2} < \frac{27}{64} is$

$$\begin{gathered} {\left( \frac{3}{4}\right)}^{6x+10-x^2} < \frac{27}{64} \\ \Rightarrow 6x+10-x^2>3 \\ \therefore x^2-6x-7<0 \\ \Rightarrow (x+1)(x-7)<0 \\ \Rightarrow x=-1,0,1,2,3,4,5,6 \end{gathered}$$