Trying to solve $|2x-15| = -x^2 – 5x -8$

Question

My first instinct was to take the positive and negative of the right hand side, resulting in
$2x-15 = -x^2 – 5x – 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers using the quadratic equation, and the second being two imaginary solutions. The problem is that, when graphed, these 2 graphs do not intersect at all, so there should be no real solutions. As for the given complex solutions, neither were considered by wolfram alpha.
Knowing that there are no real solutions, I'm confused as to how I might get complex solutions through means not already attempted and explained above.

Answer 1

The "is $2x - 15$ positive or negative" dichotomy doesn't work in $\Bbb{C}$. What we can say is that, if $-x^2 - 5x - 8 = |2x - 15|$, then $-x^2 - 5x - 8$ must be a non-negative real number, let's call it $k$. Let's start by solving $$-x^2 - 5x - 8 = k$$ for $x \in \Bbb{C}$. Using the quadratic formula, $$x = \frac{-5 \pm \sqrt{5^2 - 4(8 + k)}}{2} = \frac{-5 \pm i\sqrt{7 + 4k}}{2}.$$ Next, we need $|2x - 15|$ to be equal to this $k$. In particular, $$k = \left|2\frac{-5 \pm i\sqrt{7 + 4k}}{2} - 15\right| = \left|-20 \pm i\sqrt{7 + 4k}\right| = \sqrt{407 + 4k}.$$ Thus, we get a real equation: $$k^2 - 4k - 407 = 0, \quad k \ge 0.$$ This has a unique solution (bearing in mind $k \ge 0$): $$k = 2 + \sqrt{411}.$$ Thus, our solutions for $x \in \Bbb{C}$ must be: $$x = \frac{-5 \pm i\sqrt{7 + 4k}}{2} = \frac{-5 \pm i\sqrt{15 + 4\sqrt{411}}}{2}.$$