Is the Assumption for Equation 6.27 in Srednicki's QFT Implied?

[MathematicalPhysicist]

Homework Statement

I am reading the solutions manual of Srednicki's textbook in QFT (alongside reading the textbook itself).
Here it is:
http://www.scribd.com/doc/87916496/Srednicki-Ms-Qft-Solutions-Rev

So the equation that he arrives at (6.27).

I am not sure I understand how did he arrive at the RHS?

I mean if I write the exponent down, i.e e^{-(q_2-q_1)^2/(2c)} e^{-(q_1-q_0)^2/(2c)}
then I get:
-[(q_2-q_1)^2/2c + (q_1 -q_0)^2/2c] = -[\frac{(q_2-q_0)^2}{4c} +\frac{2q_1^2-2q_1(q_2+q_0)+q_2q_0}{2c}

Now to get the RHS we need to assume that: q_2q_0 - 2q_1(q_2+q_0)=0 which I don't see it written in the textbook, perhaps I skipped over it...

Is it written in the book?

Homework Equations

The Attempt at a Solution

 

[Oxvillian]

I got the answer in (6.27) without any extra assumptions.

Do your Gaussian integral more carefully!

 

[MathematicalPhysicist]

OK, I can see my mistake, it shouild be:

-[(q_2-q_1)^2/2x +(q_1-q_0)^2/2c] = -(q_2-q_0)^2/4c - (q_1 - (q_0+q_2)/2)^2/c

So all is OK.
:-D foolish me.