Решите уравнение: x^2-12x+36+|x^2-4x-12|=0.

\[(x - 6)^{2} + \left| x^{2} - 4x - 12 \right| = 0\]

\[(x - 6)^{2} \geq 0;\ \ \]

\[\left| x^{2} - 4x - 12 \right| \geq 0 \Longrightarrow сумма\ \]

\[равна\ 0,\ если\ каждое\]

\[выражение\ равно\ 0.\]

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

\[x - 6 = 0\ \ \]

\[x = 6.\]

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

\[D = ( - 4)^{2} - 4 \cdot 1 \cdot ( - 12) =\]

\[= 16 + 48 = 64\]

\[x_{1} = \frac{4 + \sqrt{6}4}{2} = \frac{4 + 8}{2} = \frac{12}{2} = 6\]

\[x_{2} = \frac{4 - \sqrt{6}4}{2} = \frac{4 - 8}{2} =\]

\[= - \frac{4}{2} = - 2\]

\[Ответ:x = 6.\]