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

Jan 20, 2017

$= \sqrt{- 23} = \sqrt{23} i$

#### Explanation:

It is:

$\sqrt{{1}^{2} - 4 \cdot 3 \cdot 2}$

$= \sqrt{1 - 24}$

$= \sqrt{- 23}$

It isn't a real number.

It is the complex number $\sqrt{23} i$