332. Solve the system of equations:

$$
\left\{\begin{array}{l}
2 x^{2}-7 x y-4 y^{2}+9 x-18 y+10=0 \\
x^{2}+2 y^{2}=6
\end{array}\right.
$$
$\triangle$ Let's consider the first equation of the system as a quadratic equation in $x$. Arrange its terms in the left part in descending order of $x$:

$$
2 x^{2}+(9-7 y) x-4 y^{2}-18 y+10=0
$$

Find the discriminant of the quadratic equation:

$D=(9-7 y)^{2}-8\left(-4 y^{2}-18 y+10\right)=$

$=81-126 y+49 y^{2}+32 y^{2}+144 y-80=$

$=81 y^{2}+18 y+1=(9 y+1)^{2}$.

Now we get:

$$
x=\frac{7 y-9 \pm(9 y+1)}{4} ; \quad x_{1}=4 y-2, \quad x_{2}=-\frac{y+5}{2}
$$

Substitute these expressions for $x$ into the second equation of the system and solve the two resulting quadratic equations. Complete the solution on your own.

Answer: $(2 ; 1),(-2 ;-1),(-22 / 9 ;-1 / 9)$.