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b^2+40b-500=0
a = 1; b = 40; c = -500;
Δ = b2-4ac
Δ = 402-4·1·(-500)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3600}=60$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-60}{2*1}=\frac{-100}{2} =-50 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+60}{2*1}=\frac{20}{2} =10 $
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