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?

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 !)
