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n^2+2n-1.97=0
a = 1; b = 2; c = -1.97;
Δ = b2-4ac
Δ = 22-4·1·(-1.97)
Δ = 11.88
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{-(2)-\sqrt{11.88}}{2*1}=\frac{-2-\sqrt{11.88}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{11.88}}{2*1}=\frac{-2+\sqrt{11.88}}{2} $
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