Ta có:\(Q=2x^2-6x\)
\(Q=2\left(x^2-3x\right)\)
\(Q=2\left(x^2-2.\frac{3}{2}x+\frac{9}{4}\right)-\frac{9}{2}\)
\(Q=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)
Dấu = xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Max Q = -9/2 khi x = 3/2