$-5| 2+4x | = -32(x+3/4)- | x | + 1$ This was my attempt:
$$-5| 2+4x | = -32\left(x+\frac34\right)- | x | + 1\\
\implies|2+4x|=\frac{-32x-24- | x | + 1}{-5}\\
\implies2+4x=\pm \frac{-32x+-24- x  + 1}{-5}\\
\implies4x=\pm \frac{-33x+-23 }{-5}-2\\
\implies-20x=\pm (-33x-23)+10\\
\implies-20x= -33x-13\text{ or} -20x=33x+33\\
\implies13x= -13\text{ or} -53x=33\\
\implies x=-1\text{ or} x=-\frac{33}{53}$$
Neither of these answers works when checking.
Using a graphing calculator I get an answer of $\frac{-11}{17}$ which does check, but I don't know how to get that answer using algebraic means.

You made a mistake between the second and the third line:
$$|2+4x|=\frac{-32x-24- | x | + 1}{-5}\\
2+4x=\pm \frac{-32x-24- \color{red}{x}  + 1}{-5}\\$$ instead it should have been $$2+4x=\pm \frac{-32x-24- \color{red}{|x|}  + 1}{-5}.\\$$
Going on you get
$$-10-20x=\pm(-32x-24- |x| + 1)$$
Now you should treat the 2 cases ($x\ge0$ and $x\lt0$) differently:
\begin{cases}
x\ge0 \to -10-20x=\pm(-32x-24- x+ 1)\\
x\lt0 \to -10-20x=\pm(-32x-24+x+ 1)
\end{cases}\begin{cases}
x\ge0 \to x=-\frac{33}{13} \lor -\frac{13}{53} \text{ solutions not acceptable}\\
x\lt0 \to x=-\frac{13}{11} \lor x=-\frac{11}{17}
\end{cases}
So the final solutions are $$x=-\frac{13}{11} \lor x=-\frac{11}{17}$$
