## Algebra 2 (1st Edition)

$4\pm4\sqrt2$
$3x^2-24x-48=0\\3(x^2-8x-16)=0\\x^2-8x-16=0$ The quadratic formula says that for $ax^2+bx+x$: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Hence here $x=\frac{8\pm\sqrt{(-8)^2-4(1)(-16)}}{2(1)}=\frac{8\pm\sqrt{64+64}}{2}=\frac{8\pm\sqrt{128}}{2}=\frac{8\pm8\sqrt{2}}{2}=4\pm4\sqrt2$