How do you solve 6x^2 +12x-48=0 by completing the square?

Question

How do you solve 6x^2 +12x-48=0 by completing the square?

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Eliza 1 week 2021-10-07T14:22:46+00:00 2 Answers 0

Answers ( )

    0
    2021-10-07T14:23:55+00:00

    Answer:

    x+1

    y-0

    9

    Step-by-step explanation:

    (x+1)^2 + (y-0)^2 < 9

    on edg ^u^

    0
    2021-10-07T14:24:20+00:00

    Answer:

    x = -4  and x = 2

    Through the Completing the Square process below:

    Step-by-step explanation:

    solve 6x^2 +12x-48=0

    first divide out the 6

    x^2 + 2x – 8 = 0

    Then use the Diamond method:

    -8 = -2*4 ,   and  -2 + 4 = 2

    so

    x^2 – 2x + 4x – 8 = 0

    x(x – 2) + 4*(x -2) = 0

    (x + 4)(x – 2) = 0

    so, solve  x + 4 = 0    and  (x – 2) = 0

    x = -4  and x = 2

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45:7+7-4:2-5:5*4+35:2 =? ( )