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satellite73Best ResponseYou've already chosen the best response.0
you cannot, at least you cannot using rational numbers
 one year ago

fluffyunicornBest ResponseYou've already chosen the best response.0
how do you solve it with quadratic formula? i dont think my answer was right?
 one year ago

MonkeyballBest ResponseYou've already chosen the best response.0
\[y=0.05x^2x+1 \] \[\begin{array}{*{20}c} {x = \frac{{  b \pm \sqrt {b^2  4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array} \]
 one year ago

MonkeyballBest ResponseYou've already chosen the best response.0
in this case a=0.05 b=1 c=1
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
if it was me, i would multiply everything by 20 and write \[x^220x+20=0\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
makes life a lot easier to deal with whole numbers instead of decimals
 one year ago

fluffyunicornBest ResponseYou've already chosen the best response.0
i know the formula but after i did it it looked wrong i got 1+(sr).8/.1
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
no it is just easier if it is not clear where i got the 20 from, lets clear the decimal by multiplying everything by 100 you get \[5x^2100x+100=0\] then divide by 5 and get \[x^220x+20=0\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
you really do not want to take square roots of decimals or divide by decimals use whole numbers
 one year ago

fluffyunicornBest ResponseYou've already chosen the best response.0
alright thank you!
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
for \[x^220x+20=0\] it is probably easiest to complete the square because 20 is even i would write \[x^220x+20=0\] \[x^220x=20\] \[(x10)^2=20+100=80\] \[x10=\pm\sqrt{80}=4\sqrt{5}\] \[x=10\pm4\sqrt{5}\]
 one year ago

fluffyunicornBest ResponseYou've already chosen the best response.0
And its not factorable right?
 one year ago
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