<s>[INST] System of linear and quadratic congruences Solve the system:
$x^2+x+1\equiv 0\pmod {7}$
$2x-4\equiv 0\pmod {6}$
Completing the square for first equation gives:
$$(2x+1)^2\equiv 4\pmod {7}$$
$$y^2\equiv 4\pmod{7}\Rightarrow y_1=2,y_2=5$$
$$2x+1\equiv 2\pmod{7}\Rightarrow (2,7)=1|7\Rightarrow x=4$$
$$2x+1\equiv 5\pmod{7}\Rightarrow (2,7)=1|4\Rightarrow x=2$$
Second equation:
$$2x-4\equiv 0\pmod{6}\Rightarrow (2,6)=2|4\Rightarrow x=2,x=1$$
One particular solution of the system is $x=2$.
How to find general solution? 
 [/INST] You solved the quadratic congruence correctly, but the solutions of $2x-4\equiv 0\pmod{6}$ are $x\equiv \{2,5\}\pmod{6}$.
Divide everything by $\gcd(2,4,6)=2$: $$\iff x-2\equiv 0\pmod{3}\iff x\equiv 2\pmod{3}$$
$$\iff x\equiv \{2,5\}\pmod{6}$$
The solutions of $x^2+x+1\equiv 0\pmod{7}$ are $x\equiv \{2,4\}\pmod{7}$ (as you've found).
There are $4$ solutions of the whole system: $\begin{cases}x\equiv 2\pmod{7}\\x\equiv 2\pmod{6}\end{cases}$, $\begin{cases}x\equiv 4\pmod{7}\\x\equiv 2\pmod{6}\end{cases}$, $\begin{cases}x\equiv 2\pmod{7}\\x\equiv 5\pmod{6}\end{cases}$, $\begin{cases}x\equiv 4\pmod{7}\\x\equiv 5\pmod{6}\end{cases}$. Use Chinese Remainder Theorem to find the $4$ solutions of your system.
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