Question

\[6t ^{-2}-29t ^{-1}+20=0\]

Answer 1

gimmick is to solve this as a quadratic equation is
\[t^{-1}\]

Answer 2

in other words solve
\[6x^2-29x+20=0\]
\[(x-4)(6x-5)=0\]
\[x=4,x=\frac{5}{6}\]
\[t^{-1}=4,t=\frac{1}{4}\]
\[t^{-1}=\frac{6}{5}\]
\[t=\frac{5}{6}\]

Answer 3

Thank you very much..