Find all solutions of congruence $3x^2−2x+9≡0\pmod {35}$ Find all solutions of congruence $3x^2 - 2x + 9 ≡ 0 \bmod 35$:
Attempt:
\begin{align}
3x^2 - 2x + 9 &\equiv 0 \bmod 35\tag{* 3} \\
9x^2-6x+27 &\equiv 0 \bmod 35 \tag{- 26} \\
(3x-1)^2 &\equiv -26 \bmod 35 \\
\\
-26 + 35 &= 9 = 3^2 \\
\\
\iff (3x-1)^2 &\equiv 3^2 \bmod 35 \\
\iff (3x-1-3)*(3x-1+3) &\equiv 0 \bmod 35 \\
\\
\implies \underbrace{3x - 4 \equiv 0 \bmod 35}_{(a)} 
      &\lor \underbrace{3x + 2 \equiv 0 \bmod 35}_{(b)}\\
\end{align}
Case $(a){:}\; 3x - 4 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv 4 \bmod 35$ $\Rightarrow 3x \equiv 39 \bmod 35$
$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 13 \bmod 35$
Case $(b){:}\; 3x +2 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv -2 \bmod 35$ $\Rightarrow 3x \equiv 33 \bmod 35$
$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 11 \bmod 35$
So $x = 13$ or $x = 11$.
Is it correct that way?
 A: As other comments and solutions have noted, you need to consider that 35 is composite.  But nearly all of your work is correct and completing the square is a great strategy for arbitrary quadratic congruences.
$$(3x-4)(3x+2)\equiv0\pmod{35}$$
This is correct.  But then you need to go from there to
$$(3x-4)(3x+2)\equiv0\pmod 5 \qquad\text{ and }\qquad (3x-4)(3x+2)\equiv0\pmod 7$$
we'll tackle those one at a time.
$$3x-4\equiv0\pmod 5\qquad\text{ or }\qquad 3x+2\equiv0\pmod 5\\
3x\equiv4\pmod 5\qquad\text{ or }\qquad 3x\equiv3\pmod 5\\
x\equiv3\pmod 5\qquad\text{ or }\qquad x\equiv1\pmod 5\\
$$
$$3x-4\equiv0\pmod 7\qquad\text{ or }\qquad 3x+2\equiv0\pmod 7\\
3x\equiv4\pmod 7\qquad\text{ or }\qquad 3x\equiv5\pmod 7\\
x\equiv6\pmod 7\qquad\text{ or }\qquad x\equiv4\pmod 7\\
$$
Pairing those solutions together with the CRT (Chinese Remainder Theorem) gives you the four solutions others have mentioned.
A: At no point in your proof did you split the congruence into $\bmod5$ and $\bmod7$, so you missed two solutions as I will show below:
$$3x^2-2x+4=0\bmod5\implies x^2+x+3\equiv0\bmod5\implies x\equiv\{1,3\}\bmod5$$
$$3x^2-2x+2\equiv0\bmod7\implies x^2+4x+3\equiv0\bmod7\implies x\equiv\{-1,-3\}\bmod7$$
Combining with the Chinese remainder theorem this gives $x\equiv\{6,11,13,18\}\bmod35$.
A: There is a direct way to solve modulo $35$ with the traditional quadratics method.
$\Delta^2=b^2-4ac=2^2-4\times3\times 9=-104\equiv 1\pmod {35}$
However as you can see it requires you to know how to solve $\Delta^2\equiv 1\pmod{35}$, but let assume you have a precomputed table of the squares then:
$\Delta\in\{1,6,29,34\}=\{-1,1,-6,6\}$
Then use $x=(-b\pm\Delta)\times (2a)^{-1}\pmod {35}$
$(2a)^{-1}\equiv 6^{-1}\equiv 6\pmod{35}$ and we do not have to worry about the $\pm$ in the formula since $\Delta$ already covers additive inverses.
Therefore for each $4$ values of $\Delta$ we get same number values of $x\in\{6,11,13,18\}$
