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n^2-5n-300=0
a = 1; b = -5; c = -300;
Δ = b2-4ac
Δ = -52-4·1·(-300)
Δ = 1225
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{1225}=35$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-35}{2*1}=\frac{-30}{2} =-15 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+35}{2*1}=\frac{40}{2} =20 $
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