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20y^2-13y+2=0
a = 20; b = -13; c = +2;
Δ = b2-4ac
Δ = -132-4·20·2
Δ = 9
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{9}=3$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*20}=\frac{10}{40} =1/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*20}=\frac{16}{40} =2/5 $
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