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n^2+201n-5400=0
a = 1; b = 201; c = -5400;
Δ = b2-4ac
Δ = 2012-4·1·(-5400)
Δ = 62001
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{62001}=249$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(201)-249}{2*1}=\frac{-450}{2} =-225 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(201)+249}{2*1}=\frac{48}{2} =24 $
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