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z^2+3+4*i=0 equation

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Numerical solution:

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The solution

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z  + 3 + 4*I = 0
(z2+3)+4i=0\left(z^{2} + 3\right) + 4 i = 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:
z1=D−b2az_{1} = \frac{\sqrt{D} - b}{2 a}
z2=−D−b2az_{2} = \frac{- \sqrt{D} - b}{2 a}
where D = b^2 - 4*a*c - it is the discriminant.
Because
a=1a = 1
b=0b = 0
c=3+4ic = 3 + 4 i
, then
D = b^2 - 4 * a * c = 

(0)^2 - 4 * (1) * (3 + 4*i) = -12 - 16*i

The equation has two roots.
z1 = (-b + sqrt(D)) / (2*a)

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

or
z1=1−2iz_{1} = 1 - 2 i
z2=−1+2iz_{2} = -1 + 2 i
Vieta's Theorem
it is reduced quadratic equation
pz+q+z2=0p z + q + z^{2} = 0
where
p=bap = \frac{b}{a}
p=0p = 0
q=caq = \frac{c}{a}
q=3+4iq = 3 + 4 i
Vieta Formulas
z1+z2=−pz_{1} + z_{2} = - p
z1z2=qz_{1} z_{2} = q
z1+z2=0z_{1} + z_{2} = 0
z1z2=3+4iz_{1} z_{2} = 3 + 4 i
The graph
Sum and product of roots [src] 
sum
-1 + 2*I + 1 - 2*I
(1−2i)+(−1+2i)\left(1 - 2 i\right) + \left(-1 + 2 i\right) 
=
0
00 
product
(-1 + 2*I)*(1 - 2*I)
(−1+2i)(1−2i)\left(-1 + 2 i\right) \left(1 - 2 i\right) 
=
3 + 4*I
3+4i3 + 4 i 
3 + 4*i
Rapid solution [src] 
z1 = -1 + 2*I
z1=−1+2iz_{1} = -1 + 2 i 
z2 = 1 - 2*I
z2=1−2iz_{2} = 1 - 2 i 
z2 = 1 - 2*i
Numerical answer [src] 
z1 = -1.0 + 2.0*i
z2 = 1.0 - 2.0*i
z2 = 1.0 - 2.0*i
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