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w^2+w-420=0
a = 1; b = 1; c = -420;
Δ = b2-4ac
Δ = 12-4·1·(-420)
Δ = 1681
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{1681}=41$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-41}{2*1}=\frac{-42}{2} =-21 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+41}{2*1}=\frac{40}{2} =20 $
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