We talked about this before, remember? It's a maximisation/minimisation problem. You have d^2 as a quadratic and you want to find the minimum point of d^2. There are two ways of finding the minimum point of a quadratic, either by differentiation or by completing the square.
Anyhow, you have
5t2−30t+50=5(t2−6t+10)=5((t−3)2+1)=5(t−3)2+5. Do you still remember how to complete the square?
You have
t2−6t+10 so you do
(t−26)2 but that gets you
t2−6t+9 so you need to add one.
Hence you have
t2−6t+10=(t−3)2+1.
Hence, you have:
5(t2−6t+10)=5(t−3)2+5Now, if you remember your C2/C1/Whatever, you should remember that this means your minimum point is at
(3,5) - if you don't, then I strongly recommend you go look up completing the square in your textbook.
So the minimum value of
d2 occurs when
t=3 and the minimum value
is d2=5.
That's a bit like me saying
y=x2+1 has a minimum at
(0,1) - i.e: the minimum occurs when
x=0 and the minimum value is
y=1. Except, in this case,
d2 is your
y and
x is your
t.