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20y^2-41y+20=0
a = 20; b = -41; c = +20;
Δ = b2-4ac
Δ = -412-4·20·20
Δ = 81
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{81}=9$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-9}{2*20}=\frac{32}{40} =4/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+9}{2*20}=\frac{50}{40} =1+1/4 $
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