The smallest negative integer satisfying both the quadratic inequalities  $x^2<4x+77$ and $x^2>4$  is 

$\begin{array}{rl} & x^2-4x-77<0\text{and}{x}^2-4>0\\ & \Rightarrow (x+7)(x-11)<0\text{and}(x-(-2\left)\right)(x-2)>0\\ & x\in (-7,11) and x\in (-\infty ,-2)\cup (2,\infty )\\ & \Rightarrow \text{common solution is}(-7,-2)\cup (2,11)\\ & \therefore \text{smaller negative integer}=-6\end{array}$