Question: > 
> Solve the following system of equations:
>  $$\large
> \left\{
> \begin{align\*}
> 3x^2 + xy - 4x + 2y - 2 = 0\\
> x(x + 1) + y(y + 1) = 4
> \end{align\*}
> \right.
> $$
> 
> 
> 

I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.

Answer: 
Solving the first equation for $y$ we get $$y=\frac{-3x^2+4x+2}{2+x}$$ for $$x\neq -2$$
plugging this in the second equation we get after simplifications
$$(5x+4)(x-1)^3=0$$
The resultant of $3\,{x}^{2}-xy-4\,x+2\,y-2$ and $
x \left( x+1 \right) +y \left( y+1 \right) -4$ with respect to $y$ is
$$ 10\,{x}^{4}-24\,{x}^{3}-10\,{x}^{2}+42\,x-8$$
which is irreducible over the rationals. Its roots can be written in terms of
radicals, but they are far from pleasant. There are two real roots, 
$x \approx -1.287147510$ and $x \approx 0.2049816008$, which correspond to 
$y \approx -2.469872787$ and $y \approx 1.500750095$ respectively.

EDIT: For the corrected question, the resultant of $3\,{x}^{2}+xy-4\,x+2\,y-2$ and $x \left( x+1 \right) +y \left( y+1 \right) -4$ with respect to $y$ is
$$ 10 x^4 - 22 x^3 + 6 x^2 + 14 x - 8 = 2 (5 x + 4) (x-1)^3$$
Thus we want $x = -4/5$, corresponding to $y = -13/5$, or $x = 1$, corresponding to $y = 1$.