Maximum of d(12-d)

Question

I'm a little confused on a quite simple quadratic problem. I need to calculate the maximum of $d(12-d)$ using basic quadratics. The answer is $6$ as can also be shown by $f'(x)= -2d +12$, however this is algebra not calculus. Using the squares method I get twelve, I think I may be using the incorrect method. Could someone set me on the right track?

$d(12-d)$
$= -d^2 + 12d$
$-d^2 + 12/(-2)d + 36 = 36 $
$(d-6)^2 = 36$
$d = 12 or 0$ <- I understand these are the roots, am I not meant to calculate them?

The suggested answer is: $-(d-6)^2 + 36$, but I have no idea where it came from.

Answer 1

$$d(12-d)=12d-d^2=-(d^2-2\cdot6\cdot d+6^2)+6^2=36-(d-6)^2$$

Now for real $d, (d-6)^2\ge 0$

$\implies d(12-d)\le 36$

Answer 2

You can use your knowledge of parabolas. The parabola $y=x(12-x)$ opens downwards, and reaches its maximum halfway between the two points where it crosses the $x$-axis. These two points can be read off from the equation.