3y^2+18y+24=0

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Solution for 3y^2+18y+24=0 equation: 



3y^2+18y+24=0

a = 3; b = 18; c = +24;
Δ = b2-4ac
Δ = 182-4·3·24
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*3}=\frac{-24}{6}
=-4
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*3}=\frac{-12}{6}
=-2
$

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