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

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Answer 1 · 2016-02-23T17:43:29

The solutions are:
$x= (2+4i)/2$

$x= (2-4i)/2$

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

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:

$Delta=b^2-4*a*c$

$= (-2)^2-(4*1*5)$

$= 4 -20$

$=-16$

The solutions are found using the formula:
$x=(-b+-sqrtDelta)/(2*a)$

$x = (-(-2)+-sqrt(-16))/(2*1) = (2+-sqrt(-16))/2$

$= (2+-sqrt(-1 xx 16))/2$

$= (2+-4(i))/2$

The solutions are:
$x= (2+4i)/2$

$x= (2-4i)/2$

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