# How do you find all the real and complex roots of 3x^2 - x + 2 = 0?

Aug 8, 2018

$x = \frac{1 + \sqrt{23} \cdot i}{6} \mathmr{and} x = \frac{1 - \sqrt{23} \cdot i}{6}$

#### Explanation:

Here ,

$3 {x}^{2} - x + 2 = 0$

Comparing with $a {x}^{2} + b x + c = 0$

$a = 3 , b = - 1 \mathmr{and} c = 2$

$\therefore \Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - 4 \left(3\right) \left(2\right)$

:.color(red)(Delta=1-24=-23 < 0=>ul(x !inRR and x inCC)

$\therefore \Delta = 23 \left(- 1\right)$

$\therefore \Delta = 23 {i}^{2} \to \left[\because {i}^{2} = - 1\right]$

$\therefore \sqrt{\Delta} = \sqrt{23} \cdot i$

So,

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{1 \pm \sqrt{23} \cdot i}{2 \times 3}$

$\therefore x = \frac{1 \pm \sqrt{23} \cdot i}{6}$

Hence ,

$x = \frac{1 + \sqrt{23} \cdot i}{6} \mathmr{and} x = \frac{1 - \sqrt{23} \cdot i}{6}$