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  • 11-06-2020
  • Mathematics
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how many real-number solutions does 4x^2+2x+5=0 have

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Ares123893
Ares123893 Ares123893
  • 11-06-2020

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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