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

Aug 28, 2016

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

#### Explanation:

For the equation $a {x}^{2} + b x + c = 0$, the roots are given by $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

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

In the equation ${x}^{2} - 4 x + 5 = 0$, 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)

= $\frac{4 \pm 2 i}{2}$ i.e.

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