Resposta :

Eriivan
1)

x²+4x=0
x(x+4)=0
x'→ 0
x''=-4

2 )

x²-16x=0
x(x-16)=0
x'→ 0
x''→ 16

3 )

[tex]x^2-8x+12=0\\\Delta=(-8)^2-4.1.12\\\Delta=64-48\\\Delta=16[/tex]

[tex]x= \frac{8\pm4}{2} \\ \\x_1=6\\ \\x_2=2[/tex]

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