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3y^2+9y+6=0
a = 3; b = 9; c = +6;
Δ = b2-4ac
Δ = 92-4·3·6
Δ = 9
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{9}=3$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3}{2*3}=\frac{-12}{6} =-2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3}{2*3}=\frac{-6}{6} =-1 $
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