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?

Just because we can write a question doesn't mean the question has any solutions.  For example I can with a perfectly straight face ask you to solve $|x-3|=-7$ (which is impossible as $|x-3| \ge 0 > -7$) and if you solve by squaring both sides to get $(x-3)^2 = (-7)^2$ and you do the work ($x^2-6x + 9 = 49; x^2-6x-40=(x-10)(x+4)=0$ so $x=10$ or $x=-4$) you get only extraneous solutions.
To show there is no solution we can look at the conditions.  For $\sqrt{2x+9}$ to exist we must have $2x +9 \ge $ or $x \ge -4\frac 12$.  But we must also have $\sqrt{2x+9}\ge 0$ and so $-6\sqrt{2x+9} \le 0$ and must have $x+22 < 0$ so $x < -22$.  A contradiction.
....
What about complex numbers....
The issue isn't about taking the square root of a negative number.  It's about defining which of the two square roots $2x+9$ we are going to consider to be the square root.  The only way for $x+22 = -6\sqrt{2x+9}$ is if we allow our selves have either square root be valid.
If we do that we get your extraneous roots.
