Решите уравнение х^2+12у-4х=-у^2-40.

\[x^{2} + 12y - 4x = - y^{2} - 40\]

\[x^{2} + y^{2} + 12y - 4x + 40 = 0\]

\[(x - 2)^{2} + (y + 6)^{2} = 0\]

\[x - 2 = 0\ \ \ \ \ \ \ \ \ \ y + 6 = 0\]

\[x = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = - 6\]

\[Ответ:x = 2;\ y = - 6.\ \]