+171 $ 4t^2\le 9t-2 \ \ \ \Rightarrow \ \ \ 4t^2+2\le 9t \ \ \ \Rightarrow \ \ \ 4t^2+2-9t\le \:0 $
a trinomial
$a=4$ ; $b=-9 $ ; $c=2$
plug that into $ _2t_1=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $:
$_2t_1=\frac{-\left(-9\right)\pm \sqrt{\left(-9\right)^2-4\cdot \:4\cdot \:2}}{2\cdot \:4}$
$_2t_1=\frac{-\left(-9\right)\pm \sqrt{81-32}}{2\cdot \:4}$
$_2t_1= =\frac{-\left(-9\right)\pm \:7}{2\cdot \:4}$
$t_1=\frac{9+7}{2\cdot \:4}$ and $t_2=\frac{9-7}{2\cdot \:4}$
those, respectively give you $ t_1=2 $ and $ t_2=\frac{1}{4} $
thus, we get $ \boxed{\frac{1}{4}\le t\le 2} $
