\(\text{Ta có:}A+2=x^2+3x+\frac{1}{4}+2=x^2+3x+\frac{9}{4}=x^2+2.\frac{3}{2}x+\left(\frac{3}{2}\right)^2=\left(x+\frac{3}{2}\right)^2\ge0\)
\(\Rightarrow A+2\ge0\Rightarrow A\ge-2\)
Dấu "=" xảy ra khi: \(x+\frac{3}{2}=0\Leftrightarrow x=\frac{-3}{2}\)