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    qwertyuioop's Avatar
    qwertyuioop Posts: 87, Reputation: 1
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    #1

    Jun 14, 2012, 05:50 AM
    how do i do this step by step?
    solve x^2-10x+7=0
    what I have so far is this..
    x^2-10x____=-7
    I don't know what 2 do next!
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
    Expert
     
    #2

    Jun 14, 2012, 06:14 AM
    Are you familiar with the quadratic equation? If , then

    qwertyuioop's Avatar
    qwertyuioop Posts: 87, Reputation: 1
    Junior Member
     
    #3

    Jun 14, 2012, 06:18 AM
    yes
    so then I plug it in and..
    x= -10x -+ squre root of 10x^2-4ax^2+7
    over 2(-10)
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Jun 14, 2012, 07:33 AM
    Not quite. Given x^2-10x + 7=0, the coefficient of the x^2 term is 'a,' so a=1. The coefficient of the x term is 'b,' so b=-10. And the constant at the end is c, so c=7. Plug these values into the quadratic equation to find values for x.
    qwertyuioop's Avatar
    qwertyuioop Posts: 87, Reputation: 1
    Junior Member
     
    #5

    Jun 14, 2012, 07:36 AM
    I don't get that!
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Jun 14, 2012, 07:54 AM
    I'll do a different equation for you as an example. Suppose you want to find values of x that satisfy 2x^2-4x-6=0. In this equation a=2, b=-4, and c=-6. So the quadratic equation becomes:



    So the answer is x equals either 3 or -1. To check that this is correct it's a good idea to plug these two values back into the original equation and see if it works:

    For x=3:
    2(3)^2 -4(3)-6 = 18-12-6 = 0

    For x=-1:
    2(-1)^2 -4(-1) -6 = 2+4-6 = 0

    So these values of x do indeed satisfy the original equation.

    You can apply this same process to your problem. Hope this helps.
    qwertyuioop's Avatar
    qwertyuioop Posts: 87, Reputation: 1
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    #7

    Jun 14, 2012, 07:57 AM
    Yes it does thanks! <3

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