Решите уравнение: m^4=21m^2+100.

\[m^{4} = 21m^{2} + 100\]

\[m^{4} - 21m^{2} - 100 = 0\]

\[m^{2} = x \geq 0:\]

\[x^{2} - 21x - 100 = 0\]

\[D = 441 + 400 = 841 = 29^{2}\]

\[x_{1} = \frac{21 - 29}{2} = - \frac{8}{2} < 0;\]

\[x_{2} = \frac{21 + 29}{2} = \frac{50}{2} = 25.\]

\[m^{2} = 25\]

\[m = \pm 5.\]

\[Ответ:m = \pm 5.\]