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30y^2+50y=0
a = 30; b = 50; c = 0;
Δ = b2-4ac
Δ = 502-4·30·0
Δ = 2500
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{2500}=50$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-50}{2*30}=\frac{-100}{60} =-1+2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+50}{2*30}=\frac{0}{60} =0 $
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