Решите уравнение: x^2-14x+40=0.

\[x² - 14x + 40 = 0\]

\[x = 1,\ \ b = - 14,\ \ c = 40\]

\[D = b^{2} - 4ac =\]

\[= ( - 14)^{2} - 4 \cdot 1 \cdot 40 =\]

\[= 196 - 160 = 36\]

\[x_{1,2} = \frac{- b \pm \sqrt{D}}{2a}\]

\[x_{1} = \frac{- ( - 14) + \sqrt{36}}{2 \cdot 1} = \frac{14 + 6}{2} =\]

\[= \frac{20}{2} = 10\]

\[x_{2} = \frac{- ( - 14) - \sqrt{36}}{2 \cdot 1} = \frac{14 - 6}{2} =\]

\[= \frac{8}{2} = 4\]

\[Ответ:x = 10;x = 4.\]