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20w-15w^2=0
a = -15; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-15)·0
Δ = 400
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{400}=20$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-15}=\frac{-40}{-30} =1+1/3 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-15}=\frac{0}{-30} =0 $
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