$B(x) = x^2+3x-5$
$ = x^2+2.x.\dfrac{3}{2}+\dfrac{3^2}{2^2} - \dfrac{29}{4}$
$ = \bigg(x+\dfrac{3}{2}\bigg)^2 - \dfrac{29}{4} ≥ - \dfrac{29}{4}$
Dấu "=" xảy ra $⇔x+\dfrac{3}{2} = 0 $
$\to x=\dfrac{-3}{2}$
Vậy $B(x)_{min} =\dfrac{-29}{4}$ tại $x=\dfrac{-3}{2}$
