Respostas

2013-11-14T15:52:11-02:00
Resolvendo por Bhaskara, temos:

$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
$$

$$
x=\frac{-8\pm\sqrt{8^2-4*1*16}}{2}.
$$

$$
x=\frac{-8\pm\sqrt{64-64}}{2}.
$$

$$
x=\frac{-8\pm\0}{2}.
$$

x'= -4
x''=-4