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Mathematics, 17.10.2019 14:10 maritzamartinnez

Can someone check this?
2x^2 + 50 = βˆ’20x
2x^2 + 20x + 50 = 0
2(x^2 + 10x + 25) = 0
2(x + 5)(x + 5) = 0
(x + 5)(x + 5) = 0

x + 5 = 0 or x + 5 = 0
x = βˆ’ 5, x = βˆ’ 5
the solution set is
{βˆ’5}.

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Answers: 2

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You know the right answer?
Can someone check this?
2x^2 + 50 = βˆ’20x
2x^2 + 20x + 50 = 0
2(x^2 + 10x + 25)...
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