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z^2-6*z+10=0

z^2-6*z+10=0 equation

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Numerical solution:

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The solution

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 2               
z  - 6*z + 10 = 0
$$\left(z^{2} - 6 z\right) + 10 = 0$$
Simplify
Detail solution
This equation is of the form
a*z^2 + b*z + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$z_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -6$$
$$c = 10$$
, then
D = b^2 - 4 * a * c = 

(-6)^2 - 4 * (1) * (10) = -4

Because D<0, then the equation
has no real roots,
but complex roots is exists.
z1 = (-b + sqrt(D)) / (2*a)

z2 = (-b - sqrt(D)) / (2*a)

or
$$z_{1} = 3 + i$$
$$z_{2} = 3 - i$$
Vieta's Theorem
it is reduced quadratic equation
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -6$$
$$q = \frac{c}{a}$$
$$q = 10$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = 6$$
$$z_{1} z_{2} = 10$$
The graph
Plot the graph f1(x)
Plot the graph f2(x)
Sum and product of roots [src] 
sum
3 - I + 3 + I
$$\left(3 - i\right) + \left(3 + i\right)$$ 
=
6
$$6$$ 
product
(3 - I)*(3 + I)
$$\left(3 - i\right) \left(3 + i\right)$$ 
=
10
$$10$$ 
10
Rapid solution [src] 
z1 = 3 - I
$$z_{1} = 3 - i$$ 
z2 = 3 + I
$$z_{2} = 3 + i$$ 
z2 = 3 + i
Numerical answer [src] 
z1 = 3.0 + 1.0*i
z2 = 3.0 - 1.0*i
z2 = 3.0 - 1.0*i
The graph
z^2-6*z+10=0 equation
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