Find $B=(x,y)$ so triangle is equilateral Let $O=(0,0)$, $A=(3,4)$ and $B=(x,y)$ be three points in $xOy$. Find real numbers $x$ and $y$, so that $OAB$ is an equilateral triangle.
I'm really struggling with this one, can someone help?

We need to have $|\overline{OB}|=|\overline{AB}|=|\overline{OA}|=5$. Thus from
$$ |\overline{OB}|^2=x^2+y^2=25$$
and $$|\overline{AB}|^2=(x-3)^2+(y-4)^2=25$$we reach
$$x^2-6x+9+y^2-12y+16=x^2+y^2 $$
or $$6x+12y=25\text{ or }x=2y-25/2.$$
Replace this value to $x^2+y^2=25$:
$$4y^2+50y+625/4+y^2=25$$
or $$y^2+10y-\frac{150}{4}=0 .$$
The solutions of this equation are
$$y=\frac{-1\pm5\sqrt{10}}{2} $$
The rest is obvious.
