Найду корень уравнения: x^3+y^2
a*y^2 + b*y + c = 0
D = b^2 - 4 * a * c =
(0)^2 - 4 * (1) * (x^3) = -4*x^3
y1 = (-b + sqrt(D)) / (2*a)
y2 = (-b - sqrt(D)) / (2*a)
___________________________________________________________ ___________________________________________________________
/ 2 2 / / 3 2 3 2 \\ / 2 2 / / 3 2 3 2 \\
4 / / 3 2 \ / 3 2 \ |atan2\im (x) - 3*re (x)*im(x), - re (x) + 3*im (x)*re(x)/| 4 / / 3 2 \ / 3 2 \ |atan2\im (x) - 3*re (x)*im(x), - re (x) + 3*im (x)*re(x)/|
y1 = - \/ \im (x) - 3*re (x)*im(x)/ + \- re (x) + 3*im (x)*re(x)/ *cos|---------------------------------------------------------| - I*\/ \im (x) - 3*re (x)*im(x)/ + \- re (x) + 3*im (x)*re(x)/ *sin|---------------------------------------------------------|
\ 2 / \ 2 / ___________________________________________________________ ___________________________________________________________
/ 2 2 / / 3 2 3 2 \\ / 2 2 / / 3 2 3 2 \\
4 / / 3 2 \ / 3 2 \ |atan2\im (x) - 3*re (x)*im(x), - re (x) + 3*im (x)*re(x)/| 4 / / 3 2 \ / 3 2 \ |atan2\im (x) - 3*re (x)*im(x), - re (x) + 3*im (x)*re(x)/|
y2 = \/ \im (x) - 3*re (x)*im(x)/ + \- re (x) + 3*im (x)*re(x)/ *cos|---------------------------------------------------------| + I*\/ \im (x) - 3*re (x)*im(x)/ + \- re (x) + 3*im (x)*re(x)/ *sin|---------------------------------------------------------|
\ 2 / \ 2 /