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23 January, 18:21

How many real numbers solutions are there to the equation?

0 = - 3x^2+x-4

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  1. 23 January, 19:21
    0
    ax^2+bx+c=0

    The quadratic formula is

    x = (-b + / - √ (b^2-4ac)) / 2a

    The part √ (b^2-4ac) tells us how many solution we will get:

    using - 3x^2+x-4

    √ (1-4 (-4) (-3)) = √ (1-36) = √-35 this is an impossible result since the inside of a square root has to be always greater or equal to zero. This means that the equation has no solutions.
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