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?
 A: Since $|x| = x$ or $-x$, $$|x^2+2x+2| = |x^2-3x-4|$$ if and only if
$$x^2+2x+2 = x^2-3x-4$$ or 
$$-(x^2+2x+2) = x^2-3x-4$$ or
$$x^2+2x+2 = -(x^2-3x-4)$$ or
$$-(x^2+2x+2) = -(x^2-3x-4)$$
Some cases are equivalent (can you guess which and why?), so the computations are easier than they look.
A: 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
$$
x^2+2x+2 = 
\left\{\begin{array}{lcl}-x^2+3x+4 & \rightarrow & x = \{\frac{1\pm\sqrt{17}}{4}\}\\
x^2-3x-4 & \rightarrow & x = -\frac{6}{5}\end{array}\right.
$$
A: Obviously $|x^2-3x-4|$ is greater than or equal to $0$, because you take the absolute value. This observation tells you nothing. In stead, distinguish cases where the quadratics are positive or negative.
A: There's a simply fact about absolute values (of real numbers $a$ and $b$) that you should know and understand:
$$|a|=|b| \quad \text{if and only if} \quad a=\pm b.$$
Think about it for a moment, and it should become clear to you.
Then apply this observation to the given equation: you will get two equations, as two possible cases, without absolute values, which will be easy enough to solve.
