\(E=2x^2+3x+\dfrac{9}{8}+\dfrac{55}{8}=2\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{55}{8}=2\left[x^2+2\cdot\dfrac{3}{4}x+\left(\dfrac{3}{4}\right)^2\right]+\dfrac{55}{8}=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{55}{8}\\ \left(x+\dfrac{3}{4}\right)^2\ge0\Rightarrow E\ge\dfrac{55}{8}\)
Vậy $E$ luôn dương với mọi giá trị của biến