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

For complex numbers, the expression $ \ |2x - 15| \ $ is interpreted as the modulus of the number(s) in brackets.  If we use $ \ z \ = \ a + bi \ \ , \ \ a \ , \ b \ $ real, the given equation becomes
$$  \sqrt{(2a - 15)^2 \ + \ (2b)^2 } \ \ = \ \ -(a + bi)^2 \ - \ 5(a + bi) \ - \ 8 $$
$$ \Rightarrow \ \  \sqrt{4a^2 - 60a + 225 + 4b^2 } \ \ = \ \ -(a^2 - b^2 + 5a +  8)  \ - \  (2ab + 5b)i \ \ . $$
Since the modulus is a real number, the imaginary part of the expression on the right side of this equation must be zero.  So we have $ \ 2ab + 5b \ = \ 0 \ \Rightarrow \ a \ = \ -\frac52 \ \ \text{or} \ \ b \ = \ 0 \ \ . $
Since it is easy to show that the "downward-opening" parabola $ \ y \ = \ -x^2 - 5x - 8 \ = \ -\left(x + \frac52 \right)^2 - \frac74 \ \ $ has its vertex "below" the $ \ x-$axis, there can be no purely real solutions to the original equation (as you also found from a graph).
Inserting $ \ a \ = \ -\frac52 \ \ $ into the equation instead leads to
$$  \ \sqrt{25 + 150 + 225 + 4b^2 } \ \ = \ \ -(\frac{25}{4} - b^2 - \frac{25}{2} +  8) \ \ \Rightarrow  \ \ 2·\sqrt{100 +  b^2 } \ \ = \ \ b^2 -  \frac{7}{4} $$
$$ \Rightarrow  \ \ b^4 \ - \ \frac{15}{2}·b^2 \ - \ \frac{6351}{16} \ \ = \ \ 0 \ \ , $$
a biquadratic equation with   two real solutions for $ \ b \ $ .  The two complex roots of the original equation are thus $ \ z \ = \ -\frac52 \ \pm \ i·\frac{ \sqrt{15 \ + \ 4 \sqrt{411}}}{2} \ \ . $
