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x^2-4x+5=0

x^2-4x+5=0 equation

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Numerical solution:

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The solution

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 2              
x  - 4*x + 5 = 0
$$\left(x^{2} - 4 x\right) + 5 = 0$$
Simplify
Detail solution
This equation is of the form
a*x^2 + b*x + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -4$$
$$c = 5$$
, then
D = b^2 - 4 * a * c = 

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

Because D<0, then the equation
has no real roots,
but complex roots is exists.
x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = 2 + i$$
$$x_{2} = 2 - i$$
Vieta's Theorem
it is reduced quadratic equation
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -4$$
$$q = \frac{c}{a}$$
$$q = 5$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 4$$
$$x_{1} x_{2} = 5$$
The graph
Plot the graph f1(x)
Plot the graph f2(x)
Sum and product of roots [src] 
sum
2 - I + 2 + I
$$\left(2 - i\right) + \left(2 + i\right)$$ 
=
4
$$4$$ 
product
(2 - I)*(2 + I)
$$\left(2 - i\right) \left(2 + i\right)$$ 
=
5
$$5$$ 
5
Rapid solution [src] 
x1 = 2 - I
$$x_{1} = 2 - i$$ 
x2 = 2 + I
$$x_{2} = 2 + i$$ 
x2 = 2 + i
Numerical answer [src] 
x1 = 2.0 - 1.0*i
x2 = 2.0 + 1.0*i
x2 = 2.0 + 1.0*i
The graph
x^2-4x+5=0 equation
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