$x_1=\frac{-b-\sqrt{b^2+24c}}{4}<3,\quad x_2=\frac{-b+\sqrt{b^2+24c}}{4}<3$
$x_1=\frac{-b-12-\sqrt{b^2+24c}}{4}<0,\quad x_2=\frac{-b-12+\sqrt{b^2+24c}}{4}<0$
Ilk kok herzaman negatiftir. Dolayisiyla butun $(b,c)$ ikilileri saglar.
Ikinci kok icin $ x_2=\frac{-b-12+\sqrt{b^2+24c}}{4}<0\implies -b-12+\sqrt{b^2+24c}<0$
$\sqrt{b^2+24c}<b+12\implies(\sqrt{b^2+24c})^2<(b+12)^2\implies b^2+24c<b^2+24b+144$
$\implies c-b-6<0$
$c=1: \quad b=\{1,2,3,4,5,6,7,8,9,10\} $
$c=2: \quad b=\{1,2,3,4,5,6,7,8,9,10\} $
$c=3: \quad b=\{1,2,3,4,5,6,7,8,9,10\} $
$c=4: \quad b=\{1,2,3,4,5,6,7,8,9,10\} $
$c=5: \quad b=\{1,2,3,4,5,6,7,8,9,10\} $
$c=6: \quad b=\{1,2,3,4,5,6,7,8,9,10\} $
$c=7: \quad b=\{2,3,4,5,6,7,8,9,10\} $
$c=8: \quad b=\{3,4,5,6,7,8,9,10\} $
$c=9: \quad b=\{4,5,6,7,8,9,10\} $
$c=10: \quad b=\{5,6,7,8,9,10\} $
Cevab $C)\, 90$ olacak..