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20y^2+20y=0
a = 20; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·20·0
Δ = 400
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{400}=20$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*20}=\frac{-40}{40} =-1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*20}=\frac{0}{40} =0 $
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