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

User Or Duan
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1 Answer

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Answer: Two solutions

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

User Letronje
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