Find the solutions of the next congruence using Chinese Remainder Theorem Find the solutions of the congruence using Chinese Remainder Theorem:
$2x^2 - 3x -2 \equiv 0\mod21$
By now I've done this:
$$
\left\{ 
\begin{array}{c}
2x^2-3x-2\equiv 0\mod 7\\2x^2-3x-2\equiv 0\mod3
\end{array}
\right.
$$
$$
2x^2-3x-2\equiv 0\mod 7 \\ \Delta = 25 = 5^2 \\x_1,x_2 = (2a)^{-1}*(-b \pm \sqrt\Delta) \\ x_1=4^{-1}*2 = 2*2 = 4 \mod 7 \\ x_2 = 4^{-1}*(-8)= 6*2 = 5 \mod 7
$$
Then I made $\Delta$ for the (mod 3) equation and got $x_3=1 \mod 3, x_4=2 \mod 3$
So I got:
$$
x \equiv 1 \mod 3 \\ x \equiv 2 \mod 3 \\ x \equiv 4 \mod 7 \\ x \equiv 5 \mod 7 \\ 
$$
And I don't know what to do further. I should apply Chinese Remainder Theorem on it or it would be wrong?
Edit: Thanks for the help! The correct form was:
$$
x \equiv 1 \mod 3 \\ x \equiv 2 \mod 3 \\ x \equiv 2 \mod 7 \\ x \equiv 3 \mod 7 
$$
And I understand how to solve it.
 A: Solving the quadratic equation mod. $3$ and mod. $7$, you obtain:


*

*mod. $3$: the equation i s equivalent to $2(x^2-1)=0$, and since $2$ is a unit mod. $3$, this means $x\equiv\pm 1\bmod 3$.

*mod $7$: the discriminant is $\Delta=9+16=25\equiv 4\bmod 7$, so the roots are
$$x\equiv(3\pm 2)\cdot 4^{-1}\equiv (3\pm 2)\cdot 2\equiv 2,3\mod 7.$$
There remains to combine these solutions in 4 pairs $(\alpha,\beta)$. Modulo $21$, you find the corresponding values via the inverse isomorphism of the Chinese remainder  theorem. Explicitly, start from a Bézout's relation between $7$ and $3$, say $\;7-2\cdot 3=1$. Then
$$\begin{cases}
x\equiv \alpha\mod 3,\\
x\equiv \beta\mod 7
\end{cases}\iff x\equiv\alpha \cdot 7-2\beta \cdot3\mod 21.$$
A: As suggested by saulspatz in a comment to the question,
we have $(x-2)(2x+1)\equiv0\pmod{21}$.  
That means $x\equiv 2\pmod{21}$ or $x\equiv-\dfrac12\equiv10\pmod{21}$ 
or $x\equiv2\pmod7$ and $x\equiv-\dfrac12\equiv1\pmod3$ [which means $x\equiv16\pmod{21}$]
or $x\equiv-\dfrac12\equiv3\pmod7$ and $x\equiv2\pmod3$ [which means $x\equiv17\pmod{21}].$
