Another, trivially different, way of looking at it: Let $\alpha = 2 + \sqrt3$. Then $\mathbf Q(\alpha) = \mathbf Q(\sqrt3)$. Quadratic extensions of $\mathbf Q$ are all Galois, and you know that the only non-trivial automorphism of $\mathbf Q(\sqrt3)$ is the one that performs $a + b\sqrt3 \mapsto a - b\sqrt3$.
I think the easiest way to see that symmetric polynomials in the conjugates are rational uses separability. More specifically, say I have an algebraic number with conjugates $\beta_1, \ldots, \beta_n$. If $f$ is a symmetric polynomial and $\sigma$ is some embedding of $\mathbf Q(f(\beta_1, \ldots, \beta_n))$ into an algebraic closure, then $\sigma$ permutes the conjugates and hence fixes $f(\beta_1, \ldots, \beta_n)$. By separability the degree of this field extension is equal to the number of such embeddings, and we've shown that there is exactly one. Hence this expression is rational.
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
Use the following substitution. $$x^2+18x+45=t$$ and we obtain $$t-2\sqrt{t}-15=0$$ ot $$(\sqrt{t}-1)^2=16$$ and since $\sqrt{t}\geq0$, we obtain $$\sqrt{t}=5,$$ $$x^2+18x+20=0$$ or $$(x+9)^2=61$$ and we get the answer: $$\{-9\pm\sqrt{61}\}$$