<s>[INST] 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?

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

 [/INST] You can factor out $2$ at the second summand. 
$$\sqrt{\underbrace{4+(y-6)^2}_{=5}}+2\cdot \sqrt{\underbrace{4+\left(\frac{y-3}2\right)^2}_{=5}}=\sqrt{5}+2\cdot\sqrt{5}$$
Now we see that  the following equations has to be true at the same time.


*

*$(y-6)^2=1 \Rightarrow y_1=7,y_2=5$

*$\left(\frac{y-3}2\right)^2=1\Rightarrow y_1=5,y_2=1$
Thus the solution is $y=...$
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