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n^2+21n-40=0
a = 1; b = 21; c = -40;
Δ = b2-4ac
Δ = 212-4·1·(-40)
Δ = 601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{601}}{2*1}=\frac{-21-\sqrt{601}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{601}}{2*1}=\frac{-21+\sqrt{601}}{2} $
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