Respostas

2013-11-17T19:07:47-02:00
4x^2+16x-9=0     ( Fórmula de Bhaskara)
Δ = b² -4ac
Δ = 16² -4 *4 * (-9)Δ = 256 + 144
Δ = 400
x' = (-b + √∆)/ 2a
(-16 + 20)/2 * 4
(-16 + 20)/8
4/8= 0,5
x'' = (-b - √∆)/ 2a
(-16 - 20)/2 * 4
(-16 - 20)/8
-36/8=-4,5

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2013-11-17T19:15:20-02:00
-4x2-16x+9=0
delta= b2- 4(a)(c)
delta= 256- 4(-4)(9)
delta= 256+ 144
delta= 400

 x= -b+/- raiz do delta/ 2(a)
x= 16+/- raiz de 400/ 2(-4)
x= 16+/-20/-8
x1= 36/-8→ 18/-4→ 9/-2
x2= -4/-8→ 1/2
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