Respostas

2013-10-04T09:28:02-03:00
Usando Bháskara, temos:
a=-2 , b=-2 e c=12
x = \frac{b+- \sqrt{b^2 - 4ac} }{2a}
x= \frac{-(-2)+- \sqrt{(-2)^2-4(-2)12} }{2(-2)}
x = \frac{2+- \sqrt{4+ 96} }{-4}
x = \frac{2+- \sqrt{100} }{-4}
x' = \frac{2+ 10}{-4} => \frac{12}{-4} => -3
x'' = \frac{2- 10 }{-4} => \frac{-8}{-4} => 2

1 5 1