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3y^2+29y+18=0
a = 3; b = 29; c = +18;
Δ = b2-4ac
Δ = 292-4·3·18
Δ = 625
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{625}=25$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-25}{2*3}=\frac{-54}{6} =-9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+25}{2*3}=\frac{-4}{6} =-2/3 $
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