Ta có: \(D=2x^2-8x-10\)
\(=2\left(x^2-4x-5\right)\)
\(=2\left(x^2-4x+4-9\right)\)
\(=2\left(x-2\right)^2-18\ge-18\forall x\)
Dấu '=' xảy ra khi x=2
\(D=2x^2-8x-10=2\left(x^2-4x+4\right)-8-10=2\left(x-2\right)^2-18\)
Do \(2\left(x-2\right)^2\ge0\)
\(\Rightarrow D=2\left(x-2\right)^2-18\ge-18\)
\(minD=-18\Leftrightarrow2\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)