`B = 2x^2 + 6x - 5`
`= (2x^2 + 6x) - 5`
`= 2(x^2 + 2.(3)/(4)x + 9/16 - 9/16) - 5`
`= 2(x + 3/4)^2 - 9/8 - 5`
`= 2(x + 3/4)^2 - 49/8`
Ta có:
`2(x + 3/4)^2 >= 0` vói `AA x in RR`
`=> B >= -49/8` với `AA x in RR`
Dấu "=" xảy ra:
`<=> x + 3/4 = 0`
`<=> x = -3/4`
Vậy `B_{min} = -49/8` khi `x = -3/4`