Another way to solve $| x^2 - 4x + 3 | + | x^2 - 6x + 8 | =\sqrt{4|x|^2 - 20x + 25}$ Let $| x^2 - 4x + 3 | + | x^2 - 6x + 8 | =\sqrt{4|x|^2 - 20x + 25}$ then number of integral values of $|x|$ is 
My work 
$=
\begin{cases}
2x^2 - 10x + 11= -2x + 5,  & x\ge 1\\
-2x+5 = -2x+5 , & 1 \lt x \le 2
\\
-2x^2 + 10x - 11= -2x + 5 ,& 2 \lt x \le \frac{5 }{2}\\
-2x^2 + 10x - 11= 2x - 5 ,& \frac{5}{2} \lt x \le 3 \\
2x - 5=2x -  5, & 3 \lt x \le 4
\\
2x^2 - 10x + 11= 2x - 5 ,& x \gt4
\end{cases}$
I want to know that how this question can be done in another way .

Note that $|x|^2 = x^2$ for all $x\in \mathbb R$. Using @juantheron's comment, notice that
$$|x^2-4x+3-(x^2-6x+8)| = |2x-5| = \sqrt{4x^2-20x+25}.$$
Finally, note that
$$(x^2-4x+3)(x^2-6x+8) = (x-1)(x-2)(x-3)(x-4).$$
So the set of solutions is $[1,2]\cup[3,4]$.
