Finding the intersection of $x= y^2 + y - 2$ and $y = -x^2 - \frac32x + 1 $ Goodmorning, I am struggling in finding the points of intersection of the following parabolas:
$$\begin{align}
x &= y^2 + y - 2 \\[4pt]
y &= -x^2 - \frac32x + 1
\end{align}$$
I know that these two can be solved either algebraically or with matrices, but at the moment the algebraic solution is the one that I'm going for.
How can I solve this system without getting lost in complex and boring calculations? (I'm asking specifically if there are some tricks to solve this more quickly end elegantly)

$$x = y^2 + y - 2 \tag 1$$
$$y = -x^2 - \frac32x + 1 \tag 2$$
Inspection :
$x=0$ in Eq.(2) gives $y=1$ and $y=1$ in Eq.(1) gives $x=0$ . Thus a first solution is
$$(x=0\:;\:y=1)$$
$y=0$ in Eq.(1) gives $x=-2$ and $x=-2$ in Eq.(2) gives $y=0$ . Thus a second solution is
$$(x=-2\:;\:y=0)$$
Full solving :
Put $y$ from Eq.(2) into Eq.(1) :
$$x = (-x^2 - \frac32x + 1)^2 + (-x^2 - \frac32x + 1) - 2 $$
$$x^4+3x^3-\frac34 x^2-\frac{11}{2}x=0$$
Knowing the two above roots it is easy to factor :
$$x(x+2)(4x^2+4x-11)=0$$
Solving $4x^2+4x-11)=0$ leads to the two other roots. for each one Eq.(1) gives $y$.
The third solution is :
$$x=-\frac12+\sqrt{3}\quad ; \quad y=-\frac32-\frac12\sqrt{3}$$
The fourth solution is :
$$x=-\frac12-\sqrt{3}\quad ; \quad y=-\frac32+\frac12\sqrt{3}$$
