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.

Note that $3 \equiv 5^2$, therefore $3x^2 +2x +1 \equiv 5^2x^2 + 2 \cdot 5 \cdot 5^{-1} x + 5^{-2} - 5^{-2} +1$ and since $5^{-1} \equiv 9$ this means $3x^2 +2x +1 \equiv 5^2x^2 + 2 \cdot 5 \cdot 9 x + 9^2 - 3 \equiv (5x + 9)^2 - 3$, so you want to solve $(5x + 9)^2 \equiv 3 \equiv 5^2$, therefore $(5x + 9) \equiv \pm 5$, giving $5x \equiv 7$ and $5x \equiv 8$, giving $x \equiv 8$ and $x \equiv 6$.
