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w^2+50w-1400=0
a = 1; b = 50; c = -1400;
Δ = b2-4ac
Δ = 502-4·1·(-1400)
Δ = 8100
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{8100}=90$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-90}{2*1}=\frac{-140}{2} =-70 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+90}{2*1}=\frac{40}{2} =20 $
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