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How do I solve this, please?

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2x^2 - 3x - 20 = 0

Update:

Merci, i know i was being a bit stupid but because it was 2x I kept putting (2x+5)(x-2) for some crazy reason.

8 Answers

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  • 1 decade ago
    Favourite answer

    Gotta factorise it:

    ( 2x + 5 ) ( x - 4 )

    So

    X = -2.5 & X = 4

  • 1 decade ago

    This is how we solve Quadratic Equation:

    2x^2-3x-20=0

    2x^2-8x+5x-20=0

    2x(x-4)+5(x-4)=0

    (2x+5)(x-4)=0

    x=-5/2,4

  • Anonymous
    1 decade ago

    (2x+5)(x-4)=0 so x=-5/2 or x=4

  • 1 decade ago

    2x squared -3x -20=0 +20

    2x squared -3x = 20 i then used elimination

    x=1

    2-3=-1

    x=2

    8-6= 2

    x=3

    18-9=9

    x=4

    32-12=20

    so thats your answer

    x=4

  • >_<
    Lv 4
    1 decade ago

    2x^2-3x-20=0

    (2x+5)(x-4)

    2x+5=0

    -5+5=0

    2x=-5

    x=-2.5 or -5/2

    Now for the x-4=0

    4-4=0

    x=4

    So the answers are x=-2.5 or -5/2

    and x=4

    Source(s): I just studied this in AP Algebra 2!!!
  • Anonymous
    1 decade ago

    enter it into a graphing calculator in one of the y= spots

    in the other enter in y=0

    then intersect the two points

    orr you could do the quadratic equation..

    b plus or minus the square root of (b squared minus 4(a)(c)) all divided by 2a.

    so 2= a, b= -3 and c=-20

    -3 plus or minus the square root of (-3squared minus 4*2*-20) all divided by 2*2

  • 1 decade ago

    I know its mean, but I would help you if you actually showed you had tried the question and had some sort of idea of an answer :o)

  • 1 decade ago

    you can use the quadratic formula

    or

    factorise it

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