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20s^2-19s+3=0
a = 20; b = -19; c = +3;
Δ = b2-4ac
Δ = -192-4·20·3
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*20}=\frac{8}{40} =1/5 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*20}=\frac{30}{40} =3/4 $
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