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20w^2+3w-9=0
a = 20; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·20·(-9)
Δ = 729
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{729}=27$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*20}=\frac{-30}{40} =-3/4 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*20}=\frac{24}{40} =3/5 $
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