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26 May, 05:22

How many real-number solutions does 4x^2+2x+5=0 have

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  1. 26 May, 05:36
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    Answer: Two solutions

    Step-by-step explanation:

    1. x = (-2-√-76) / -8 = (1+i√ 19) / 4 = 0.2500-1.0897i

    2. x = (-2+√-76) / -8 = (1-i√ 19) / 4 = 0.2500+1.0897i
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