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

Question

asked Jul 2, 2021 52.8k views

1 Answer

Letronje · Answer 1 · 2021-07-08T14:35:14+0000

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