Answer
$t=1$
Work Step by Step
$\frac{3}{t}-\frac{t^2-2t}{t^3} = \frac{4}{t^2}$
$t^3*(\frac{3}{t}-\frac{t^2-2t}{t^3} = \frac{4}{t^2})$
$\frac{3t^3}{t}-\frac{(t^2-2t)(t^3)}{t^3} = \frac{4t^3}{t^2}$
$3t^2 -(t^2-2t) = 4t$
$3t^2 -t^2+2t = 4t$
$2t^2+2t=4t$
$2t^2+2t-4t=4t-4t$
$2t^2-2t = 0$
$2t(t-1)=0$
$2t=0$
$t=0$ (we can't have zero in the denominator, so this answer is invalid)
$t-1=0$
$t=1$
$t=1$
$\frac{3}{t}-\frac{t^2-2t}{t^3} = \frac{4}{t^2}$
$\frac{3}{1}-\frac{1^2-2*1}{1^3} = \frac{4}{1^2}$
$3-(1-2)/1 = 4$
$3-(-1)/1 = 4$
$3--1 = 4$
$4=4$
