The number of integral values of x satisfying the inequality ${\left(\frac{3}{4}\right)}^{6\mathrm{x}+10-{\mathrm{x}}^2}<\frac{27}{64}$ is __________.

${\left(\frac{3}{4}\right)}^{6\mathrm{x}+10-{\mathrm{x}}^2}<\frac{27}{64}$

$\begin{array}{l}\text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\mathrm{x}+10-{\mathrm{x}}^2>3\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}^2-6\mathrm{x}-7<0\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{x}+1)(\mathrm{x}-7)<0\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}=0,1,2,3,4,5,6\end{array}$