Generally speaking, the problem arises because squaring is not a "reversible" operation. That is, while it is true that if $a=b$ then $a^2=b^2$, it is not true that if $a^2=b^2$ then $a=b$. (For instance, even though $(-1)^2=1^2$, it does not follow that $-1=1$)
This is in contrast to other kinds of equation manipulations that we use routinely when we solve equations. For example, if $a=b$, then $a+k=b+k$, and conversely: if $a+k=b+k$, then $a=b$. So we can add to both sides of an equation (for instance, you can go from $\sqrt{x+5}+1 = x$ to $\sqrt{x+5}=x-1$ by adding $-1$ to both sides) without changing the solution set of the equation. Likewise, we can multiply both sides of an equation by a nonzero number, because $a=b$ is true if and only if $ka=kb$ is true when $k\neq 0$. We can also take exponentials (since $a=b$ if and only if $e^a=e^b$) and so on.
But squaring doesn't work like that, because it cannot be "reversed". If you try to reverse the squaring, you run into a rather big problem; namely, that $\sqrt{x^2}=|x|$, and is not equal to $x$.
So when you go from $\sqrt{x+5} = x-1$ to $(\sqrt{x+5})^2 = (x-1)^2$, you are considering a new problem. Anything that was a solution to the old problem ($\sqrt{x+5}=x-1$) is still a solution to the new one, but there may be (and in fact are) things that are solutions to the new problem that do not solve the old problem.
Any such solutions (solutions to the new problem that are not solutions to the original problem) are sometimes called "extraneous solutions". Extraneous means "coming from the outside". In this case, it's a solution that comes from "outside" the original 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
Don't rush to square equations. Instead, first try to eliminate surds via substitution.
Let $y=\sqrt{x+2}$. Then the equation becomes
$$2x^2 - 7xy + 3y^2=0$$ Which you can factor as a quadratic $$(2x-y)(x-3y)=0$$ Or (using $x=y^2-2$) $$(2y^2-y-4)(y^2-3y-2)=0$$ Find the largest root of each quadratic, then the overall largest root, then get back to $x$.