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5n^2+9n-2000=0
a = 5; b = 9; c = -2000;
Δ = b2-4ac
Δ = 92-4·5·(-2000)
Δ = 40081
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{-(9)-\sqrt{40081}}{2*5}=\frac{-9-\sqrt{40081}}{10} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{40081}}{2*5}=\frac{-9+\sqrt{40081}}{10} $
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