jkev1
Use the quadratic formula to find the zeros of y=2x^2+8x+1



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ParthKohli
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\[x = {b \pm \sqrt{b^2  4ac} \over 2a}\]

jkev1
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thats not one of the options @ParthKohli

jkev1
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@Coolsector

jkev1
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@yahoo!

Coolsector
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you should use what parthkohli wrote ..
ax^2+bx+c =0
use what he wrote

jkev1
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it's not an option though @Coolsector

Coolsector
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no .. look
ax^2+bx+c =0
in your case
a=2
b =8
c=1
now plug it into what parthkoli wrote and you will get the zeros ..

jkev1
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Look at this.

jkev1
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@Coolsector

Coolsector
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if you plug it into what partkholi wrote you get :
\[\frac{ 8 \pm \sqrt{648} }{ 4 }\]