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2x^2-8x+11=0

2x^2-8x+11=0 equation

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Numerical solution:

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The solution

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   2               
2*x  - 8*x + 11 = 0
$$\left(2 x^{2} - 8 x\right) + 11 = 0$$
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
$$a = 2$$
$$b = -8$$
$$c = 11$$
, then
D = b^2 - 4 * a * c = 

(-8)^2 - 4 * (2) * (11) = -24

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
$$x_{1} = 2 + \frac{\sqrt{6} i}{2}$$
$$x_{2} = 2 - \frac{\sqrt{6} i}{2}$$
Vieta's Theorem
rewrite the equation
$$\left(2 x^{2} - 8 x\right) + 11 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 4 x + \frac{11}{2} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -4$$
$$q = \frac{c}{a}$$
$$q = \frac{11}{2}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 4$$
$$x_{1} x_{2} = \frac{11}{2}$$
The graph
Rapid solution [src]
             ___
         I*\/ 6 
x1 = 2 - -------
            2   
$$x_{1} = 2 - \frac{\sqrt{6} i}{2}$$
             ___
         I*\/ 6 
x2 = 2 + -------
            2   
$$x_{2} = 2 + \frac{\sqrt{6} i}{2}$$
x2 = 2 + sqrt(6)*i/2
Sum and product of roots [src]
sum
        ___           ___
    I*\/ 6        I*\/ 6 
2 - ------- + 2 + -------
       2             2   
$$\left(2 - \frac{\sqrt{6} i}{2}\right) + \left(2 + \frac{\sqrt{6} i}{2}\right)$$
=
4
$$4$$
product
/        ___\ /        ___\
|    I*\/ 6 | |    I*\/ 6 |
|2 - -------|*|2 + -------|
\       2   / \       2   /
$$\left(2 - \frac{\sqrt{6} i}{2}\right) \left(2 + \frac{\sqrt{6} i}{2}\right)$$
=
11/2
$$\frac{11}{2}$$
11/2
Numerical answer [src]
x1 = 2.0 - 1.22474487139159*i
x2 = 2.0 + 1.22474487139159*i
x2 = 2.0 + 1.22474487139159*i
The graph
2x^2-8x+11=0 equation