Ta có: \(3x^2+2x-5=3\left(x^2+\frac{2}{3}x-\frac{5}{3}\right)\)
\(=3\left(x^2+2.\frac{1}{3}x+\frac{1}{9}-\frac{16}{9}\right)\)
\(=3\left[\left(x+\frac{1}{3}\right)^2-\frac{16}{9}\right]\)
\(=3\left(x+\frac{1}{3}\right)^2-\frac{16}{3}\ge\frac{-16}{3}\left(????\right)\)