The smallest negative integer satisfying both the quadratic...

Adding~$4$~on both sides

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

\left(x - 2\right)2 < 81$</p><p>$- 9 < x - 2 < 9$</p><p>$- 7 < x < 11 . . . . . . . . . . . . . . . . . . . . . . . . \left(i\right)$</p><p>Now,~$x^{2} > 4$</p><p>$x^{2} - 4 > 0$</p><p>$x > 2$or~$x < - 2 . . . . . . . . . . . . . . . . . . \left(ii\right)</p><p>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 $- 3.</p>$