3y2+6y-9=0

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Solution for 3y2+6y-9=0 equation: 



3y^2+6y-9=0

a = 3; b = 6; c = -9;
Δ = b2-4ac
Δ = 62-4·3·(-9)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12}{2*3}=\frac{-18}{6}
=-3
$


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

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