Stuck, please help, anyone

midpoint of AB  \(=\,(\frac{t-4+-2}{2}\,,\,\frac{-1+t+3}{2})\,=\,(\frac{t-6}{2}\,,\,\frac{t+2}{2})\)

an endpoint of AB  \(=\,(t-4,-1)\)

distance between midpoint and an endpoint  \(=\,\ \,\sqrt{(t-4-\frac{t-6}{2})^2+(-1-\frac{t+2}{2})^2}\)

\(\begin{array} \ {(t-4-\frac{t-6}{2})^2+(-1-\frac{t+2}{2})^2}\,&=&\,\frac{t^2}{2} \\~\\ {(\frac{2t}{2}-\frac{8}{2}-\frac{t-6}{2})^2+(-\frac{2}{2}-\frac{t+2}{2})^2}\,&=&\,\frac{t^2}{2} \\~\\ {(\frac{2t-8-t+6}{2})^2+(\frac{-2-t-2}{2})^2}\,&=&\,\frac{t^2}{2} \\~\\ {(\frac{t-2}{2})^2+(\frac{-t-4}{2})^2}\,&=&\,\frac{t^2}{2} \\~\\ {(\frac{t-2}{2})(\frac{t-2}2)+(\frac{-t-4}{2})(\frac{-t-4}{2})}\,&=&\,\frac{t^2}{2} \\~\\ {\frac{t^2-4t+4}{4}+\frac{t^2+8t+16}{4}} \,&=&\,\frac{t^2}{2} \\~\\ {\frac{2t^2+4t+20}{4}}\,&=&\,\frac{t^2}{2} \\~\\ 2t^2+4t+20 \,&=&\,2t^2 \\~\\ 4t+20\,&=&\,0 \\~\\ 4t\,&=&\,-20 \\~\\ t\,&=&\,-5 \end{array}\)

Let A,B  be the points on the coordinate plane with coordinates \(\tbinom{t-4}{-1}\)  and \(\tbinom{-2}{t+3}\), respectively.
The square of the distance between the midpoint of  \( \overline{AB}\) and an endpoint of   \(\overline{AB}\) is equal to  \(\tfrac{t^2}{2}\).
What is the value of \(t\) ?

\(\begin{array}{|rcll|} \hline \left ( \frac{ \overline{AB} } {2} \right)^2 &=& \frac{t^2}{2} \\ \frac{ \overline{AB}^2 } {4} &=& \frac{t^2}{2} \\ \mathbf{ \overline{AB}^2 } & \mathbf{=} & \mathbf{2 t^2 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \overline{AB}^2 &=& [~ (t-4)-(-2) ~ ]^2 + [~ (-1) - (t+3) ~ ]^2 \\ \overline{AB}^2 &=& (~ t-4+2 ~)^2 + (~ -1 - t -3 ~ )^2 \\ \overline{AB}^2 &=& (~ t-2 ~)^2 + (~ - t -4 ~ )^2 \\ \overline{AB}^2 &=& (~ t-2 ~)^2 + [~ - (t +4) ~ ]^2 \\ \overline{AB}^2 &=& (~ t-2 ~)^2 + (~ t +4 ~)^2 \quad & | \quad \overline{AB}^2 = 2t^2\\ 2t^2 &=& (~ t-2 ~)^2 + (~ t +4 ~)^2 \\ 2t^2 &=& t^2 -4t + 4 + t^2 + 8t + 16 \\ 0 &=& -4t + 4 + 8t + 16 \\ 0 &=& 4t + 20 \\ 4t &= -20 \\ t &=& -\frac{20}{4} \\ \mathbf{ t } & \mathbf{=} & \mathbf{-5} \\ \hline \end{array}\)