3y2+3y-18=0

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Solution for 3y2+3y-18=0 equation: 



3y^2+3y-18=0

a = 3; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·3·(-18)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-15}{2*3}=\frac{-18}{6}
=-3
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+15}{2*3}=\frac{12}{6}
=2
$

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