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500n^2+1500n-11000=0
a = 500; b = 1500; c = -11000;
Δ = b2-4ac
Δ = 15002-4·500·(-11000)
Δ = 24250000
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24250000}=\sqrt{250000*97}=\sqrt{250000}*\sqrt{97}=500\sqrt{97}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1500)-500\sqrt{97}}{2*500}=\frac{-1500-500\sqrt{97}}{1000} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1500)+500\sqrt{97}}{2*500}=\frac{-1500+500\sqrt{97}}{1000} $
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