Real Values

editor keeps mangling my answer.

The final answer is

$\dfrac{t(2t-3)}{4t-2}\leq 0 \Rightarrow t \in (\infty,0] \bigcup \left(\dfrac 1 2, \dfrac 2 3\right]$

Here's one way to do this

Just  set this = 0   and   multiply through by  4t - 2   and we get that

t (3t - 2)  =  0

Setting each factor to  0    and solving, we get that 

t = 0      or     t = 2/3 

And since the denominator of the original fraction cannot = 1/2....we have these  possible intervals

(-inf, 0]  ( 0, 1/2)  (1/2, 2/3 ]  and ( 2/3, inf )

Picking a point in each interval  shows that the intervals   (-inf, 0 ]  and (1/2, 2/3]  solve the inequality

Yes! Thanks, everyone!