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n^2+31n-90=0
a = 1; b = 31; c = -90;
Δ = b2-4ac
Δ = 312-4·1·(-90)
Δ = 1321
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{-(31)-\sqrt{1321}}{2*1}=\frac{-31-\sqrt{1321}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+\sqrt{1321}}{2*1}=\frac{-31+\sqrt{1321}}{2} $
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