Решите уравнение: x^2-16x+64+|x^2-3x-40|=0.

\[x - 8 = 0\]

\[x = 8.\]

\[D = ( - 3)^{2} - 4 \cdot 1 \cdot ( - 40) =\]

\[= 9 + 160 = 169\]

\[x_{1} = \frac{3 + \sqrt{169}}{2 \cdot 1} = \frac{3 + 13}{2} =\]

\[= \frac{16}{2} = 8\]

\[x_{2} = \frac{3 - \sqrt{169}}{2 \cdot 1} = \frac{3 - 13}{2} =\]

\[= - \frac{10}{2} = - 5\]

\[Ответ:\ x = 8.\]