Find $y$ in $\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$ I would like to solve this equation for $y$. Any tips? It seems like you cant really do it analytically?

$$\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$$


Hint: After squaring one times we get
$$2\sqrt{4+(y-6)^2}\sqrt{16+(y-3)^2}=25-(y-6)^2-(y-3)^2$$
squaring again and simplfying we get
$$4 \left(46 y^2-450 y+975\right)=0$$
