# How do you solve x^2-2x+5=0 using the quadratic formula?

Feb 23, 2016

The solutions are:
$x = \frac{2 + 4 i}{2}$

$x = \frac{2 - 4 i}{2}$

#### Explanation:

${x}^{2} - 2 x + 5 = 0$

The equation is of the form color(blue)(ax^2+bx+c=0 where:

$a = 1 , b = - 2 , c = 5$

The Discriminant is given by:

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$= {\left(- 2\right)}^{2} - \left(4 \cdot 1 \cdot 5\right)$

$= 4 - 20$

$= - 16$

The solutions are found using the formula:
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$x = \frac{- \left(- 2\right) \pm \sqrt{- 16}}{2 \cdot 1} = \frac{2 \pm \sqrt{- 16}}{2}$

$= \frac{2 \pm \sqrt{- 1 \times 16}}{2}$

$= \frac{2 \pm 4 \left(i\right)}{2}$

The solutions are:
$x = \frac{2 + 4 i}{2}$

$x = \frac{2 - 4 i}{2}$