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20w^2+41w+2=0
a = 20; b = 41; c = +2;
Δ = b2-4ac
Δ = 412-4·20·2
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1521}=39$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-39}{2*20}=\frac{-80}{40} =-2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+39}{2*20}=\frac{-2}{40} =-1/20 $
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