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3

3y^(2) + 8y + 2 = 0


User Ejaenv
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1 Answer

5 votes


a {y}^(2) + by + c = 0


3 {y}^(2) + 8y + 2 = 0


a = 3


b = 8


c = 2

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∆ = {b}^(2) - 4ac


∆ = ({8})^(2) - 4 * (3) * (2)


∆ = 64 - 24


∆ = 40

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y = ( -b ± √(∆) )/(2a) \\

##############################


y(1) = ( - 8 + √(40) )/(6) \\


y(1) = ( - 8 + √(4 * 10) )/(6) \\


y(1) = ( - 8 + 2 √(10) )/(6) \\


y(1) = (2( - 4 + √(10)) )/(2 * 3) \\


y(1) = ( - 4 + √(10) )/(3) \\

+++++++++++++++++++++++++++++++++++++++


y(2) = ( - 8 - √(40) )/(6) \\


y(2) = ( - 8 - √(4 * 10) )/(6) \\


y(2) = ( - 8 - 2 √(10) )/(6) \\


y(2) = (2( - 4 - √(10)) )/(2 * 3) \\


y(2) = ( - 4 - √(10) )/(3) \\

##############################

_________________________________

And we're done.....♥️♥️♥️♥️♥️

User Leopold Stotch
by
7.9k points

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