Question

Determine the zeros of the expression -16t^2-8t+24.

Answer 1

$-16t^2-8t+24=-8(2t^2+t-3)=-8(2t^2+3t-2t-3)\\\\=-8[t(2t+3)-1(2t+3)]=-8(2t+3)(t-1)\\\\\text{the zeros}\\\\2t+3=0\to t=-1.5\\t-1=0\to t=1\\\\Answer:\ \boxed{-1.5\ and\ 1}$