Question

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

Answer 1

`=sqrt(-23)=sqrt(23)i`

Explanation:

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`