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kdsim
Group Title
Math Check:
Use the quadratic equation to solve:
2x^23x=5
I got\frac{3\pm i \sqrt{49}}{4}\
 2 years ago
 2 years ago
kdsim Group Title
Math Check: Use the quadratic equation to solve: 2x^23x=5 I got\frac{3\pm i \sqrt{49}}{4}\
 2 years ago
 2 years ago

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kdsim Group TitleBest ResponseYou've already chosen the best response.0
grr\[\frac{3\pm i \sqrt{49}}{4}\ \]
 2 years ago

Brunna Group TitleBest ResponseYou've already chosen the best response.1
2x² 3x +5 = 0 delta: (3)²  4*5 *2 = = 9  40 = 31 ~ Are you sure the equation is 2x² 3x+5 =0 not 2x²3x5 =0 ?
 2 years ago

kdsim Group TitleBest ResponseYou've already chosen the best response.0
well it was 2x^23x=5, that translates to 2x3x+50 doesn't it? :o
 2 years ago

kdsim Group TitleBest ResponseYou've already chosen the best response.0
=0 not 0 sorry
 2 years ago

Brunna Group TitleBest ResponseYou've already chosen the best response.1
so, If is 2x² 3x +5=0 delta = b² 4*a*c = 31 and \[(b \pm \sqrt{\Delta})\div 2 *a =\] \[((3)\pm \sqrt{31}) / 4 =\] \[(3\pm i \sqrt{31} )/ 4\] \[x' = (3+ i \sqrt{31}) /4\] \[x'' = (3 i \sqrt{31}) /4\]
 2 years ago
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