<s>[INST] Completing the Square : $2+0.8x-0.04x^2$ 
Write $2+0.8x-0.04x^2$  in the form $A-B(x+C)^2$, where A, B and C are constants to be determined.

Here's how I have tried it out:
$2+0.8x-0.04x^2$
$-0.04x^2+0.8x+2$
$-0.04[(x-10)^2-10]+2$
$-0.04(x-10)^2+o.8+2$
$2.8-0.04(x-10)^2$
So the answer should be, $A=2.8$ $B=0.04$ $C=-10$.
But on my book solution sheet, $A = 6$. Could you please help me out what have I done wrong?
 [/INST] Let
$$2+0.8x-0.04x^2=-\frac4{100}\left(x^2-20x-50\right)=-\frac4{100}\left(x^2-20x+100-150\right)$$
$$-\frac4{100}\left((x-10)^2-150\right)=-0.04(x^2-10)^2+6$$
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