3y2+8y+4=0

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Solution for 3y2+8y+4=0 equation: 



3y^2+8y+4=0

a = 3; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·3·4
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*3}=\frac{-12}{6}
=-2
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*3}=\frac{-4}{6}
=-2/3
$

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