Question

3

`3y^(2) + 8y + 2 = 0`
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Answer 1

`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) \\`

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`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) \\`

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And we're done.....♥️♥️♥️♥️♥️