Quick way to find solution $(\sqrt3,3)$ to the system $x^2+y^2-4y=0$ and $y=(1-\frac{\sqrt3}{3})x^2+x$? I would like to propose an intersection from an exam exercise for a Mathematical Analysis assignment at the university. This is a small part that involves calculating a double integral.

I have the following system to solve:
\begin{cases} x^2+y^2-4y=0 \\ y=(1-\frac{\sqrt3}{3})x^2+x \end{cases}
I know the points of intersection are $(0,0)$ $(\sqrt3,3)$.
I would like to be able to quickly find the point $(\sqrt3,3)$ without using Cardano's equations. Is there something I'm missing?

I have tried various approaches but I always find myself with laborious calculations that waste a lot of time. Is there any geometric approach or some parabola or circumference property to help find that point without going through a third degree polynomial? Unfortunately, the resolution of the resulting third degree polynomial is not trivial.

$y=(1-\frac{\sqrt3}{3})x^2+x$
$x^2+y^2-4y=0$
$x^2+y^2-4y+4=x^2+(y-2)^2=4$
$x^2+(y-2)^2=x^2+((1-\frac{\sqrt3}{3})x^2+x-2)^2=4$
$x^2-3+((1-\frac{\sqrt3}{3})x^2+x-2)^2-1=0$
$(x-\sqrt3)(x+\sqrt3)+((1-\frac{\sqrt3}{3})x^2+x-2-1)((1-\frac{\sqrt3}{3})x^2+x-2+1)=0$
$(x-\sqrt3)(x+\sqrt3)+\frac{1}{9}((3-\sqrt3)x^2+3x-9)((3-\sqrt3)x^2+3x-3)=0$
$(x-\sqrt3)(x+\sqrt3)+\frac{1}{9}((x-\sqrt3)((3-\sqrt3)x+3\sqrt3))((3-\sqrt3)x^2+3x-3)=0$
$(x-\sqrt3)[(x+\sqrt3)+\frac{1}{9}((3-\sqrt3)x+3\sqrt3))((3-\sqrt3)x^2+3x-3)]=0$
$x-\sqrt3=0$
$x=\sqrt3$
