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n^2-2n-2400=0
a = 1; b = -2; c = -2400;
Δ = b2-4ac
Δ = -22-4·1·(-2400)
Δ = 9604
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{9604}=98$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-98}{2*1}=\frac{-96}{2} =-48 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+98}{2*1}=\frac{100}{2} =50 $
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