Math, asked by sangeetasharma75, 4 months ago

6+4y+y^2 and 8-5y+3y^2​

Answers

Answered by pgaidhane821
1

Step-by-step explanation:

6+4y+y^2

y^2+4y+6=0

compared with ay^2+by+c=0

a=1, b=4, c=6.

b^2-4ac=4^2-4*1*6

=16-24

= - 6

this equation roots are not real numbers

8-5y+3y^2

3y^2-5y+8=0

compaired with ay^2+by+c=0

a=3, b= -5, c=8

b^2-4ac= (-5)^2-4*3*8

=25-96

= -71

this equation roots are not real numbers

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