math

Solve give your answer in interval notation: 5-4x^2>=8x!!! Please help ASAP!!!:(

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  1. 0 ≥ 4 x^2 + 8 x - 5

    0 ≥ (2 x + 5) (2 x - 1)

    for their product to be positive (>0), both factors must have the same sign

    both ≥ 0 ... x ≥ 1/2

    both ≤ 0 ... x ≤ -5/2

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  2. 5-4x^2 ≥ 8x
    -4x^2 - 8x + 5 ≥ 0
    4x^2 + 8x - 5 ≤ 0
    (2x + 5)(2x - 1) ≤ 0
    Where is the parabola y = (2x+5)(2x-1) below the x-axis ?

    -5/2 ≤ x ≤ 1/2

    https://www.wolframalpha.com/input/?i=5-4x%5E2-+8x%E2%89%A50

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