x2+3x+2=min{∣x−3∣,∣x+2∣} 
$\begin{aligned}
& y=x^2+3 x+2 \
& y=x^2+2\left(\frac{3}{2}\right) x+\frac{9}{4}-\frac{9}{4}+2 \
& y=\left(x+\frac{3}{2}\right)^2-\frac{1}{4} \
& y+\frac{1}{4}=\left(x+\frac{3}{2}\right)^2 \
& \Rightarrow \text { Parabola vertex }\left(\frac{-3}{2}, \frac{-1}{4}\right)
\end{aligned}\Rightarrow$ By graph 2 solution possible