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21 September, 04:54

Using the quadratic formula to solve 5x = 6x2 - 3, what are the values of x? StartFraction 5 plus-or-minus 3 StartRoot 11 EndRoot Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 47 EndRoot Over 12 EndFraction StartFraction negative 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction

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  1. 21 September, 08:29
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    Using the quadratic formula to solve 5x = 6x^2 - 3, what are the values of x?

    5x = 6x^2 - 3

    Subtract 5x from both sides:

    0 = 6x^2 - 5x - 3

    a = 6, b = - 5, c = - 3

    x = (-b ± √ (b^2 - 4ac)) / (2a)

    x = ( - (-5) ± √ ((-5) ^2 - 4 (6) (-3))) / (2 (6))

    x = (5 ± √ (25 + 72)) / 12

    x = (5 ± √ (97)) / 12
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