What is the solution for $x+22=-6\sqrt{2x+9}$ So, I want to solve for x in the radical equation:
$x + 22 = -6\sqrt{2x+9}$
By Squaring each expression we get:
$(x + 22)^2 = (-6\sqrt{2x+9})^2$
$ x^2 + 44x + 484 = 36\cdot(2x+ 9) $
$ x^2 + 44x + 484 = 72x + 324 $
Now by solving the quadratic equation:
$ x^2  - 28x + 160 = 0 $
$x^2 - 20x - 8x + 160 = 0$
$x\cdot(x-20) -8\cdot(x-20) = 0 $
$ (x-20)\cdot(x-8) = 0 $
$ x = 20 $ or $ x = 8 $
But, none of the values of x satisfies the equation $x + 22 = -6\sqrt{2x+9}$, they satisfy the equation $x + 22 = 6\sqrt{2x+9}$. There should be an extraneous root that satisfies the equation $x + 22 = 6\sqrt{2x+9}$, but boot the roots satisfies this equation and none of them satisfies $x + 22 = -6\sqrt{2x+9}$.
Why is that so? And is there any complex/imaginary solution to the equation?

Sometimes it can be useful to find the domain of the equation before solving equations involving radicals.
By definition of the principal square root, if $\sqrt {f(x)}=g(x)$, then $f(x)\geqslant0\wedge g(x)\geqslant 0$ holds, where $f(x)$ and $g(x)$ are some algebraic expressions.
Thus, you have:
$$
\begin{align}&\begin{cases}\sqrt {2x+9}=\frac {x+22}{-6}\geqslant 0\\ 2x+9\geqslant0\end{cases}\\\\
\implies &\begin{cases}x\leqslant-22\\ x\geqslant-\frac 92\end{cases}\\\\
\implies &x\in\color{#c00}{\emptyset.}\end{align}
$$
Therefore, we don't need to solve the equation. Because, the domain of the equation is the empty set. This means, the real solution, indeed, doesn't exist.

In this part of the answer, we want to prove that the equation $x+22=-6\sqrt{2x+9}$ has no complex roots in general.
Let $\sqrt {2x+9}=z, z\in\mathbb C\setminus \mathbb R$. Then we have:
$$
\begin{align}x+22=\frac {z^2}{2}+\frac {35}{2}=\frac {z^2+35}{2}\end{align}
$$
This leads to:
$$
\begin{align}&\frac {z^2+35}{2}=-6z\\
\implies &z^2+12z+35=0\\
\implies &\Delta_z=36-35=1\geqslant 0.\end{align}
$$
This implies that $z\in\mathbb R$, which gives a contradiction. This completes the proof.
