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hartnnBest ResponseYou've already chosen the best response.1
then the two roots of x are: \(\huge{x=\frac{b \pm \sqrt{b^24ac}}{2a}}\)
 one year ago

wioBest ResponseYou've already chosen the best response.0
Use the quadratic foumula \[x_1 =\frac{ b + \sqrt{b^24ac}} {2a}\] \[x_2 =\frac{ b  \sqrt{b^24ac}} {2a}\]
 one year ago

dddanBest ResponseYou've already chosen the best response.0
ya but i got 4+ the square root of 24 / 4 does this make sense?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
your b=5, how did u get 4+/ ... ?
 one year ago

dddanBest ResponseYou've already chosen the best response.0
sry the b=4 and the c=5
 one year ago

wioBest ResponseYou've already chosen the best response.0
You have imaginary roots. Is that what you're worried about?
 one year ago

wioBest ResponseYou've already chosen the best response.0
Why? Is it a physics problem and you are supposed to have real roots? Because there is nothing wrong with the quadratic formula. Double check your coefficients
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
u actually have imaginary roots for this..
 one year ago

dddanBest ResponseYou've already chosen the best response.0
alright can u tell me where to take it from the imaginary roots?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
how did u get square root of 24 ?
 one year ago

dddanBest ResponseYou've already chosen the best response.0
because with the coefficients, its (4)^24(2)(5) , in the square root right?
 one year ago

dddanBest ResponseYou've already chosen the best response.0
no i i mixed up the coefficents in the original question. i was given a=2;b=4;c=5
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
ok, \(\huge{x=\frac{4 \pm \sqrt{24}}{4}}=\frac{4 \pm 2\sqrt{6}}{4}=\frac{2\pm i\sqrt{6}}{2}\)
 one year ago

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