$\text{ Number of integral values of }x\text{ for which }\sqrt{8+2x-x^2}>6-3x\text{ is true is }$

$\begin{array}{l}\sqrt{8+2x-x^2}>6-3x\\ \text{ We must have }8+2x-x^2\ge 0\end{array}$

$\begin{array}{l}\Rightarrow x\in [-2,4]-----\left(1\right)\\ \text{ We must have }6-3x\ge 0\end{array}$

$\begin{array}{l}\Rightarrow x\le 2-----\left(2\right)\\ \Rightarrow 8+2x-x^2>36+9x^2-36x\\ \Rightarrow 5x^2-19x+14<0\\ \Rightarrow (5x-14)(x-1)<0\\ \Rightarrow x\in \left(1,\frac{14}{5}\right)-----\left(3\right)\end{array}$

$\begin{array}{l}\text{ by (i) and (ii) and (iii) }\\ x\in (1,4]\end{array}$