Ta có
$-2x^2 - 3x - 8 = -(2x^2 + 3x + 8) = -[(x\sqrt{2})^2 + 2 . x\sqrt{2} . \dfrac{3}{2\sqrt{2}} + (\dfrac{3}{2\sqrt{2}})^2 + \dfrac{55}{8}]$
$= -(x\sqrt{2} + \dfrac{3}{2\sqrt{2}})^2 - \dfrac{55}{8} \leq -\dfrac{55}{8}$
Dấu "=" xảy ra khi và chỉ khi $x\sqrt{2} = \dfrac{3}{2\sqrt{2}}$ hay $x = \dfrac{3}{4}$.