$\displaystyle \displaystyle \begin{align*} 5 \left[ \left( t - \frac{8}{5} \right) ^2 - \frac{64}{25} \right] &= 5 \cdot \left( t - \frac{8}{5} \right)^2 + 5 \cdot \left( -\frac{64}{25} \right) \\ &= 5\left( t - \frac{8}{5} \right)^2 - \frac{64}{5} \end{align*}$
So your turning point should be $\displaystyle \displaystyle \begin{align*} \left( \frac{8}{5}, -\frac{64}{5} \right) \end{align*}$.
No no no no, just no. That image is what the answer the solution manual has. THAT's NOT my answer. My answer is with the 1/5. I know how to expand, but no matter how I calculate it, it won't show the k value to have 1/5. Again, it is easily testable by jut plugging in k into the original polynomial above.