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3y^2+28y+9=0
a = 3; b = 28; c = +9;
Δ = b2-4ac
Δ = 282-4·3·9
Δ = 676
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{676}=26$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-26}{2*3}=\frac{-54}{6} =-9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+26}{2*3}=\frac{-2}{6} =-1/3 $
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