jmc

Let $f(t)=t(2t-3)/(4t-2).$ We make a sign table for each of the three factors on the left-hand side: \begin{array}{c|ccc|c} &$t$ &$2t-3$ &$4t-2$ &$f(t)$ \\ \hline$t<0$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$0<t<\frac{1}{2}$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$\frac{1}{2}<t<\frac{3}{2}$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$t>\frac{3}{2}$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}Therefore, we have $f(t)<0$ when $t<0$ or $\frac{1}{2}<t<\frac{3}{2}.$ Because the inequality is non-strict, we must also include the values of $t$ for which $f(t)=0,$ which are $t=0$ and $t=\frac{3}{2}.$ Putting all this together, we get that the set of solutions for $t$ is $\boxed{(-\infty ,0]\cup (\frac{1}{2},\frac{3}{2}]}.$