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n^2-2n=99
We move all terms to the left:
n^2-2n-(99)=0
a = 1; b = -2; c = -99;
Δ = b2-4ac
Δ = -22-4·1·(-99)
Δ = 400
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{400}=20$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-20}{2*1}=\frac{-18}{2} =-9 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+20}{2*1}=\frac{22}{2} =11 $
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