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hartnnBest ResponseYou've already chosen the best response.2
so, you have a quadratic equation \[x^2+2x+9=0\] where you want to solve for x ? Compare this quadratic equation with \(ax^2+bx+c=0\) find \[a=...?\\b=...?\\c=...?\] then the two roots of x are: \(\huge{x_{1,2}=\frac{b \pm \sqrt{b^24ac}}{2a}}\)
 one year ago

itsjustme_lolBest ResponseYou've already chosen the best response.1
sorry, yes, i am here. but I already solved it on my own...thankyou though :D
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
good! welcome anyways ^_^
 one year ago
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