Respostas

2013-11-16T21:55:06-02:00
2x(3x+4y-2)
6x^2+8xy-4x
6x^2-4xy=0

Δ = b² -4ac
Δ = (-4y)² -4 *6 * 0
Δ = 16y^2- 0
Δ = 16y^2
x' = (-b + √∆)/ 2a
(-(-4y) + 4y)/2 * 6
(4y + 4y)/12
8y/12= 2y/3

 x'' = (-b - √∆)/ 2a
(-(-4y) - 4y)/2 * 6
(4y - 4y)/12
0/12= 0