Solving simple mod equations Solve $3x^2 + 2x + 1 \equiv 0 \mod 11$
Additionally, I have an example problem, but a step in the middle has confused me:
$3x^2 + 5x - 7 \equiv 0 \mod 17$. Rearrange to get $3x^2 + 5x \equiv 7 \mod 17$.
$\implies 6\cdot 3x^2 + 6\cdot 5x \equiv 6*7 \mod 17$.
The next line reads $x^2 + 30x \equiv 8 \mod 17$, which I am confused about;
I understand how $8$ came to be ($42 - 34$, reduced to $8$) but what happened to the coefficient of $3x^2$? This example is from my notes in NT class, so perhaps I copied something incorrectly, or did I?
For the first problem, you don't need to finish the whole problem. Once you get to a point where it looks something like $(x+15)^2 \equiv 10 \mod 17$  (result of the example problem), I know how to proceed. It's the beginning parts that have muddled me a bit...
Help would be greatly appreciated.

$3x^2 + 2x + 1 \equiv 0 \pmod{11}$
Since $3 \cdot 4 \equiv 12 \equiv 1 \pmod{11}, \quad \frac 13 \equiv 4 \pmod{11}$
So, multiplying both sides by $4$, we get 
$x^2 + 8x + 4 \equiv 0 \pmod{11}$
Half of $8$ is $4$ and $4^2 = 16 \equiv 5 \pmod{11}$. So
$x^2 + 8x + 4 
 \equiv x^2 + 8x + 5 - 1
 \equiv (x+4)^2 - 1 \pmod{11}$
So you get 
$(x+4)^2 - 1 \equiv 0 \pmod{11}$
$(x+4)^2 \equiv 1 \pmod{11}$
$x+4 \equiv 1 \pmod{11} \quad \text{or} \quad x+4 \equiv -1 \pmod{11}$
$x \equiv 8 \pmod{11} \quad \text{or} \quad x \equiv 6 \pmod{11}$
Note that this implies
$3x^2+2x+1 \equiv 3(x-6)(x-8) \pmod{11}$
