How do you solve the quadratic with complex numbers given $x^2-4x+5=0$ ?

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Answer 1 · 2016-08-28T15:11:21

$x=2+i$ and $x=2-i)$

For the equation $ax^2+bx+c=0$ , the roots are given by $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

It is apparent that if the discriminant $b^2-4ac<0$ , we have complex roots.

In the equation $x^2-4x+5=0$ , the discriminant is $(-4)^2-4×1×5=16-20=-4<0$ and hence roots are complex.

These are $x=(-(-4)+-sqrt(-4))/(2×1)$

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

$x=2+i$ and $x=2-i$

How do you solve the quadratic with complex numbers given x^2-4x+5=0x^2-4x+5=0?