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20y^2-43y+21=0
a = 20; b = -43; c = +21;
Δ = b2-4ac
Δ = -432-4·20·21
Δ = 169
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{169}=13$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-13}{2*20}=\frac{30}{40} =3/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+13}{2*20}=\frac{56}{40} =1+2/5 $
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