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x^2+40x-41=0
a = 1; b = 40; c = -41;
Δ = b2-4ac
Δ = 402-4·1·(-41)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1764}=42$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-42}{2*1}=\frac{-82}{2} =-41 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+42}{2*1}=\frac{2}{2} =1 $
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