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?

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.
