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0.25n^2+19.75n-600=0
a = 0.25; b = 19.75; c = -600;
Δ = b2-4ac
Δ = 19.752-4·0.25·(-600)
Δ = 990.0625
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{-(19.75)-\sqrt{990.0625}}{2*0.25}=\frac{-19.75-\sqrt{990.0625}}{0.5} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19.75)+\sqrt{990.0625}}{2*0.25}=\frac{-19.75+\sqrt{990.0625}}{0.5} $
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