Basic Mathematics

$\dfrac{8x^2+16x-51}{(2x-3)(x+4)}<3\\ \dfrac{8x^2+16x-51-3(2x^2+5x-12)}{(2x-3)(x+4)}<0\\ \dfrac{2x^2+x-15}{(2x-3)(x+4)}<0\\ \dfrac{(2x-5)(x+3)}{(2x-3)(x+4)}<0$

Then, draw a number line or a table to find the range of values that satisfy this inequality:

Therefore, the range is $-4<x<-3,\;\dfrac{3}{2}<x<\dfrac{5}{2}$ or $(-4,-3)\cup\left(\dfrac{3}{2},\dfrac{5}{2}\right)$

Within this range of values of $x$ , there is only one integer: $x=2$

Therefore, the number of integral values that satisfies this inequality is $\boxed{1}$