Решите уравнение (x^2-12x+20)^2=(x^2+2x-12)^2.

\[\left( x^{2} - 12x + 20 \right)^{2} = \left( x^{2} + 2x - 12 \right)^{2}\]

\[1)\ x^{2} - 12x + 20 = x^{2} + 2x - 12\]

\[- 14x + 32 = 0\]

\[- 2(7x - 16) = 0\]

\[7x = 16\]

\[x = \frac{16}{7}.\]

\[2)\ x^{2} - 12x + 20 = - x^{2} - 2x + 12\]

\[2x^{2} - 10x - 8 = 0\ \]

\[2\left( x^{2} - 5x - 4 \right) = 0\]

\[2(x - 4)(x - 1) = 0\]

\[x = 4;\ \ x = 1\]

\[Ответ:x = \frac{16}{7};\ \ x = 1;\ \ x = 4.\]