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5n^2+1505n-12360=0
a = 5; b = 1505; c = -12360;
Δ = b2-4ac
Δ = 15052-4·5·(-12360)
Δ = 2512225
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{2512225}=1585$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1505)-1585}{2*5}=\frac{-3090}{10} =-309 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1505)+1585}{2*5}=\frac{80}{10} =8 $
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