Well the whole thing is multiplied by $-2$, meaning that it'll be largest when what's inside the bracket is smallest (since "smaller" negative numbers are in fact larger). So we want to minimize
$(t-\frac 7 4)^2+\frac{39}{16}$
By picking a certain $t$. The $\frac{39}{16}$ doesn't change with $t$, so it won't affect our minimization. We can throw it out. So we've reduced the problem to minimizing
$(t-\frac 7 4)^2$
which is simple: squares are always positive (or zero), so the smallest value this can take is zero. And it takes this value only when $t=\frac{7}{4}$. So the function takes its largest value at $t=\frac 7 4$. Plugging this into the function will give you the maximum value that the function takes.