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x^2+y^2=0 equation

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Numerical solution:

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The solution

You have entered [src] 
 2    2    
x  + y  = 0
x2+y2=0x^{2} + y^{2} = 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:
x1=D−b2ax_{1} = \frac{\sqrt{D} - b}{2 a}
x2=−D−b2ax_{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=y2c = y^{2}
, then
D = b^2 - 4 * a * c = 

(0)^2 - 4 * (1) * (y^2) = -4*y^2

The equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)

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

or
x1=−y2x_{1} = \sqrt{- y^{2}}
x2=−−y2x_{2} = - \sqrt{- y^{2}}
Vieta's Theorem
it is reduced quadratic equation
px+q+x2=0p x + q + x^{2} = 0
where
p=bap = \frac{b}{a}
p=0p = 0
q=caq = \frac{c}{a}
q=y2q = y^{2}
Vieta Formulas
x1+x2=−px_{1} + x_{2} = - p
x1x2=qx_{1} x_{2} = q
x1+x2=0x_{1} + x_{2} = 0
x1x2=y2x_{1} x_{2} = y^{2}
The graph
Rapid solution [src] 
x1 = -I*re(y) + im(y)
x1=−ire⁡(y)+im⁡(y)x_{1} = - i \operatorname{re}{\left(y\right)} + \operatorname{im}{\left(y\right)} 
x2 = -im(y) + I*re(y)
x2=ire⁡(y)−im⁡(y)x_{2} = i \operatorname{re}{\left(y\right)} - \operatorname{im}{\left(y\right)} 
x2 = i*re(y) - im(y)
Sum and product of roots [src] 
sum
-I*re(y) + im(y) + -im(y) + I*re(y)
(−ire⁡(y)+im⁡(y))+(ire⁡(y)−im⁡(y))\left(- i \operatorname{re}{\left(y\right)} + \operatorname{im}{\left(y\right)}\right) + \left(i \operatorname{re}{\left(y\right)} - \operatorname{im}{\left(y\right)}\right) 
=
0
00 
product
(-I*re(y) + im(y))*(-im(y) + I*re(y))
(−ire⁡(y)+im⁡(y))(ire⁡(y)−im⁡(y))\left(- i \operatorname{re}{\left(y\right)} + \operatorname{im}{\left(y\right)}\right) \left(i \operatorname{re}{\left(y\right)} - \operatorname{im}{\left(y\right)}\right) 
=
                   2
-(-im(y) + I*re(y))
−(ire⁡(y)−im⁡(y))2- \left(i \operatorname{re}{\left(y\right)} - \operatorname{im}{\left(y\right)}\right)^{2} 
-(-im(y) + i*re(y))^2
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