Найду корень уравнения: y^2=2cx
a*y^2 + b*y + c = 0
D = b^2 - 4 * a * c =
(0)^2 - 4 * (1) * (-2*c*x) = 8*c*x
y1 = (-b + sqrt(D)) / (2*a)
y2 = (-b - sqrt(D)) / (2*a)
_____________________ _____________________
___ 4 / 2 2 /atan2(im(c*x), re(c*x))\ ___ 4 / 2 2 /atan2(im(c*x), re(c*x))\
y1 = - \/ 2 *\/ im (c*x) + re (c*x) *cos|-----------------------| - I*\/ 2 *\/ im (c*x) + re (c*x) *sin|-----------------------|
\ 2 / \ 2 / _____________________ _____________________
___ 4 / 2 2 /atan2(im(c*x), re(c*x))\ ___ 4 / 2 2 /atan2(im(c*x), re(c*x))\
y2 = \/ 2 *\/ im (c*x) + re (c*x) *cos|-----------------------| + I*\/ 2 *\/ im (c*x) + re (c*x) *sin|-----------------------|
\ 2 / \ 2 /