Question

ayudame joseph y sebastian cortes

Answer 1

Respuesta:

$\frac{-b + \sqrt{b^{2} - 4ac} }{2a} y \frac{-b - \sqrt{b^{2} - 4ac} }{2a} \\\\=\frac{-(-13) + \sqrt{(-13)^{2} - 4(3)(-56)} }{2(3)} \\ =\frac{13 + \sqrt{169 -(-672)} }{6}\\ =\frac{13 + \sqrt{841} }{6} =\frac{13 + 29 }{6} =\frac{42}{6} =7\\\\ =\frac{13 - \sqrt{169 -(-672)} }{6}\\\\ =\frac{13 - \sqrt{841} }{6} =\frac{13 - 29}{6} =\frac{-16}{6} -\frac{8}{3}$