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3y^2+35y+50=0
a = 3; b = 35; c = +50;
Δ = b2-4ac
Δ = 352-4·3·50
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-25}{2*3}=\frac{-60}{6} =-10 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+25}{2*3}=\frac{-10}{6} =-1+2/3 $
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