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hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2so, 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: \(\huge{x_{1,2}=\frac{b \pm \sqrt{b^24ac}}{2a}}\)

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2good! welcome anyways ^_^
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