<s>[INST] Finding all values in $\mathbb R$ for quadratic absolute value equation The question:

Determine the solution set (in $\mathbb R$) for the equation $|x^2+2x+2| = |x^2-3x-4|$

So far, I have determined that for this to be true, $|x^2-3x-4|$ must be greater or equal to $0$, giving $(x-4)(x+1)\ge0$. To find the solution set, $|x^2+2x+2|\ge-1$ as indicated by the roots of the RHS equation, but this is where I get stuck. 
Where am I going wrong?
 [/INST] Knowing that $x^2+2x+2 > 0$
The equation is equivalent to
$$
x^2+2x+2 = \vert x^2-3x-4\vert
$$
which is equivalent to
$$
x^2+2x+2 = 
\left\{\begin{array}{lcl}-x^2+3x+4 & \rightarrow & x = \{\frac{1\pm\sqrt{17}}{4}\}\\
x^2-3x-4 & \rightarrow & x = -\frac{6}{5}\end{array}\right.
$$
</s>