equação do 2º grau x (x+3) - 40 = 0?

anonymousanonymous
in Science & Mathematics

Gente, me ajuda por favor?

Comments

  • polvopolvo

    x(x+3)-40 = 0

    x²+ 3x-40=0

    agora, bhaskara

    delta = b²- 4ac => 3²- 4 . 1. (-40)

    9+160 = 169

    raiz de delta = 13

    x ' = -3+ 13 /2 => 5

    x'' = -3 - 13/2 => -8

  • james-morris_60a2f2152eccdjames-morris_60a2f2152eccd

    x . (x+3) -40=0

    x² + 3x - 40 =0

    delta= (3)²-4.1.(-40)

    delta = 9+160

    delta = 169

    - ( 3)+/-V169/2.1

    -3+13/2 = 5

    -3-13/2 = -8

  • anonymousanonymous

    A partir de x (x+3) -40=0, obtemos:

    x²+3x-40=0

    △= 3²-4*1*(-40)

    △ = 9+160

    △ = 169

    x1=(-3+13)/2 <=> x1=10/2 <=> x1= 5

    x2=(-3-13)/2 <=> x2= -16/2 <=> x2= -8

    LOGO, V={-8 , 5}

  • boudicaboudica

    X² + 3 - 40 = 0

    D = b² - 4.a.c

    D= 3² - 4 . 1. (-40)

    D= 169

    X'=- 3 + 13 / 2.1

    X'= 10/2

    X'=5

    X'' = -3 -13 /2.1

    X''= -16/2

    X''-8

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