TITLE: Solve $2x^2+y^2-z=2\sqrt{4x+8y-z}-19$
QUESTION [5 upvotes]: I am trying to solve the following equation.
$$
2x^2+y^2-z=2\sqrt{4x+8y-z}-19
$$
To get rid of the square root, I tried squaring both sides which lead to
$$
(2x^2+y^2-z+19)^2=16x+32y-4z
$$
which was too complex to deal with.
Also, I have tried some substitutions to simplify the equation, but none of them were working.
I believe that the equation could be solved with a appropriate substitution and factorization, yet I have no idea what to do.
Any hint or help is appreciated.

REPLY [2 votes]: To make it a little easier to get Michael Rozenberg's answer, you can replace $4x+8y-z=t$:
$$2x^2+y^2-(4x+8y-t)=2\sqrt{t}-19 \Rightarrow \\
2(x^2-2x+1)+(y^2-8y+16)+(t-2\sqrt{t}+1)=0 \Rightarrow \\
2(x-1)^2+(y-4)^2+(\sqrt{t}-1)^2=0 \Rightarrow \\
x-1=y-4=\sqrt{t}-1=0.$$