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.

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 $x\equiv17\pmod{21}].$
