Solve $|x^2+6x+8|=|x^2+4x+5|+|2x+3|$ Question : Find solution for $x \in \Bbb R$, $|x^2+6x+8|=|x^2+4x+5|+|2x+3|$
I considered 8 different cases and arrived at the answer $\big [ \frac{-3}{2}, \infty)$
I dont know if its correct. Also, considering 8 different cases is tedious. Is there any other method to solve problems like these?
 A: The equation is piecewise quadratic and the pieces are delimited by the zeros of the arguments of the absolute values.
We have
$$|x^2+6x+8|=\begin{cases}x\le-4\lor x\ge-2&\to x^2+6x+8\\-4\le x\le-2&\to-(x^2+6x+8),\end{cases}$$
$$|x^2+4x+5|=x^2+4x+5,$$
$$|2x+3|=\begin{cases}x\le-\dfrac32&\to-(2x+3)\\x\ge-\dfrac32&\to2x+3.\end{cases}$$
There are actually four pieces to consider:
$$\begin{cases}
x\le-4&\to x^2+6x+8=x^2+4x+5-(2x+3)\to\text{incompatible} \left(x=-\dfrac32\right)\\
-4\le x\le-2&\to-(x^2+6x+8)=x^2+4x+5-(2x+3)\to\text{no roots}\\
-2\le x\le-\dfrac32&\to x^2+6x+8=x^2+4x+5-(2x+3)\to x=-\dfrac32\\
-\dfrac32\le x&\to x^2+6x+8=x^2+4x+5+2x+3\to\text{any }x.
\end{cases}$$
Note that this is a general approach. As noticed by Jaideep Khare, there is a shortcut for this particular problem.

In the case of $q$ quadratic terms with real roots and $l$ linear terms,  the combinations lead to $2q+l+1$ intervals (if you handled all bounds independently, you would have to consider $3^q2^l$ cases !)
A: If you observe the equality, it's of the form $$|u+v|=|u|+|v|$$
But we also have following triangle inequality $$|u+v| \le |u|+|v|  \quad \forall \; u,v \in \Bbb R$$
This implies that either both $u$ and $v$ are positive, or both are negative. 
Hence, only required condition is $$u \cdot v \ge 0$$
Now, since $u=x^2+4x+5=(x+2)^2+1$, it's always positive.
Therefore we need to solve for $v \ge 0$ i.e. $$(2x+3) \ge 0 \implies x \ge - \frac 32$$ 
