How do you evaluate $sqrt(b^2-4ac)$ for a=3, b=1, c=2?

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Answer 1 · 2017-01-20T23:21:55

$=sqrt(-23)=sqrt(23)i$

It is:

$sqrt(1^2-4*3*2)$

$=sqrt(1-24)$

$=sqrt(-23)$

It isn't a real number.

It is the complex number $sqrt(23)i$

How do you evaluate sqrt(b^2-4ac)sqrt(b^2-4ac) for a=3, b=1, c=2?