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

$x^2-4x<77$

Adding $4$ on both sides

$x^2-4x+4<77+4$

${\left(x-2\right)}^2<81$

$-9<x-2<9$

$-7<x<11........................\left(i\right)$

Now, $x^2>4$

$x^2-4>0$

$x>2$or $x<-2..................\left(ii\right)$

The solution set satisfying both inequalities is the common part displayed in the diagram, namely, $\left(-7,-2\right)\cup \left(2,11\right)$. Therefore, the greatest negative integer in this set is $-3$.