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10n^2+n-15=0
a = 10; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·10·(-15)
Δ = 601
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{-(1)-\sqrt{601}}{2*10}=\frac{-1-\sqrt{601}}{20} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{601}}{2*10}=\frac{-1+\sqrt{601}}{20} $
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