Explanation

[Solution by Dr. Abubacker] $|x-3|\le|x+2|$ for $x\ge\frac12$, otherwise the minimum is $|x+2|$. For $x<\frac12$: $x^2+3x+2=|x+2|$ gives $x=0$ or $x=-2$ (both valid). For $x\ge\frac12$: $x^2+3x+2=|x-3|$ gives $x^2+4x-1=0$ with $x=-2\pm\sqrt5$, neither $\ge\frac12$ (and $x^2+2x+5=0$ has no real root for $x\ge3$). So there are $2$ solutions. Answer (3).