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anonymous
 5 years ago
need help solving quadratic equation..
anonymous
 5 years ago
need help solving quadratic equation..

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0.... ..... .. .... . .. . .... .... .. . .... . (morse code)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I need to consider this problem.. \[3x ^{2}+x5=0\] And then use discriminant \[b ^{2}4ac\] and then solve with the formula \[x=b + or  \sqrt{b sqar4ac} a\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry that took so long, idk how to do the formula right on this thing.. does it kinda make sense though?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so the squared part of that is all over 2a

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0do you mean \[x=(b+\sqrt{b^24ac})/2 \] and \[x=(b\sqrt{b^24ac})/2 \]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the discriminate gives us a +number, so there are 2 solutions

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0why does it put the a so close to the x? its hard to read

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0yay i left out the time a in the denominator lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry my computer was being slow so everything you guys said just loaded.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I left off with \[1\sqrt{59} all \over 6\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i'm not sure if i do something with i at this point..or if that is even correct what I have.. if that was the answer someone gave up there I guess I'm not sure what that's supposed to look like typed out
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