$B=4x^2-x+2$
$=4(x^2-\frac{1}{4}x+\frac{1}{2})$
$=4\Big(x^2-2.x.\frac{1}{8}+\frac{1}{64}-\frac{1}{64}+\frac{1}{2}\Big)$
$=4\Big[\Big(x-\frac{1}{8}\Big)^2+\frac{31}{64}\Big]$
$=4\Big(x-\frac{1}{8}\Big)^2+\frac{31}{16}≥\frac{31}{16}$
Dấu "=" xảy ra $⇔x-\frac{1}{8}=0 ⇔x=\frac{1}{8}$
Vậy GTNN là $\frac{31}{16}$ tại $x=\frac{1}{8}$