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What are the roots of the quadratic equation below

 

 Apr 14, 2020
 #1
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Hint: Apply the quadratic formula 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

For a quadratic equation : \(ax^2+bx+c=0\)

 

 

solution:

\(x = {-15 \pm \sqrt{15^2-4*3*(-5)} \over 2(3)}\)\( = {-15 \pm \sqrt{165} \over 6}\)

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 Apr 14, 2020

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