# How do you solve the quadratic 2x^2-6x+5=0 with complex numbers?

Dec 15, 2017

$x = \frac{3}{2} \pm \frac{1}{2} i$

#### Explanation:

$0 = 2 \left(2 {x}^{2} - 6 x + 5\right)$

$\textcolor{w h i t e}{0} = 4 {x}^{2} - 12 x + 10$

$\textcolor{w h i t e}{0} = {\left(2 x\right)}^{2} - 2 \left(2 x\right) \left(3\right) + {3}^{2} + 1$

$\textcolor{w h i t e}{0} = {\left(2 x - 3\right)}^{2} + {1}^{2}$

$\textcolor{w h i t e}{0} = {\left(2 x - 3\right)}^{2} - {i}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 x - 3\right) - i\right) \left(\left(2 x - 3\right) + i\right)$

$\textcolor{w h i t e}{0} = \left(2 x - 3 - i\right) \left(2 x - 3 + i\right)$

So:

$2 x = 3 \pm i$

and:

$x = \frac{3}{2} \pm \frac{1}{2} i$