Answer
$t=1$ is the only solution.
Work Step by Step
$\frac{3}{t}-\frac{t^2-2t}{t^3}=\frac{4}{t^2}$
$\frac{3t^2-t^2+2t}{t^3}=\frac{4}{t^2}$
$\frac{2t^2+2t}{t^3}=\frac{4}{t^2}$
$2t^4+2t^3=4t^3$
$2t^4-2t^3=0$
$2t^3(t-1)$
$t=0$ or $t=1$
Check:
$\frac{3}{0}-\frac{0^2-2.0}{0^3}=\frac{4}{0^2}$
$0=0$
$\frac{3}{1}-\frac{1^2-2.1}{1^3}=\frac{4}{1^2}$
$4=4$
The solutions are $t=0$ or $t=1$.
However, $t=0$ makes the bottom of the original equation 0, so it is not a valid answer.
Thus, $t=1$ is the only solution.
