Since you have a quartic, there are potentially 4 real solutions.
Note that any solution to $x = x^2 - 3x - 2$ is also a solution to the given equation.
Hence, $x^2 - 4x - 2$ is a factor of the quartic. Now find the other factor, and solve both quadratics.
Alternatively, see this almost 10 year old thread on the Art of Problem Solving Forum, or this slightly newer thread which discusses a similar problem.
Let the unknowns be $\,a,b,c,d\,$ and divide the linear equations by $\,m_e\,$, then the system is of the form:
$$
\begin{align}
a^2 + b^2 &= r \tag 1 \\
c^2 + d^2 &= s\tag 2 \\
c + m a &= p \tag 3 \\
d + m b &= q \tag 4
\end{align}
$$
Substituting $\,c=p-ma\,$ and $\,d=q-mb\,$ from $\,(3)$-$(4)\,$ into $\,(2)\,$ gives:
$$
\begin{align}
s &= (p-ma)^2+(q-mb)^2 \\
&= p^2+q^2 - 2m(pa+qb)+m^2(\color{blue}{a^2+b^2}) \\
&= p^2+q^2 +m^2 \color{blue}{r} - 2m(pa+qb) \tag{5}
\end{align}
$$
Rearranging $\,(5)\,$:
$$
2mq\,b = p^2+q^2 +m^2r -s - 2mp\,a \;\;\iff\;\; b = \lambda - \mu a\tag{6}
$$
Substituting $\,b\,$ from $\,(6)\,$ into $\,(1)\,$ gives the quadratic in $\,a\,$:
$$
a^2 + (\lambda - \mu a)^2 = r \tag{7}
$$
Best Answer
It's $$2x^2+y^2-4x-8y+18+4x+8y-z-2\sqrt{4x+8y-z}+1=0$$ 0r $$2(x-1)^2+(y-4)^2+(\sqrt{4x+8y-z}-1)^2=0,$$ which gives $$x-1=y-4=\sqrt{4x+8y-z}-1=0.$$ Can you end it now?