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w^2+20w-1500=0
a = 1; b = 20; c = -1500;
Δ = b2-4ac
Δ = 202-4·1·(-1500)
Δ = 6400
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{6400}=80$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-80}{2*1}=\frac{-100}{2} =-50 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+80}{2*1}=\frac{60}{2} =30 $
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