Solving $\sqrt{6x-y^2-4z+1}-\sqrt{y+4z^2}=\sqrt{9x^2+y+4}$

Question

Solve over reals:

$$\sqrt{6x-y^2-4z+1}-\sqrt{y+4z^2}=\sqrt{9x^2+y+4}$$

I tried squaring

$$6x-y^2-4z+1+4z^2-2\sqrt{(6x-y^2-4z+1)(y+4z^2)}=9x^2+4$$

or

$$(2z-1)^2-y^2=3+(3x-1)^2+2\sqrt{(6x-y^2-4z+1)(y+4z^2)}$$

But this does not seem to be leading anywhere.

Answer 1

We have: $$6x-y^2-4z+1=y+4z^2+9x^2+y+4+2\sqrt{(y+4z^2)(9x^2+y+4)}$$ or $$(3x-1)^2+(y+1)^2+(2z+1)^2+2\sqrt{(y+4z^2)(9x^2+y+4)}=0.$$ Thus, we obtain the following system: $$3x-1=y+1=2z+1=(y+4z^2)(9x^2+y+4)=0.$$ Can you end it now?