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20n^2-405=0
a = 20; b = 0; c = -405;
Δ = b2-4ac
Δ = 02-4·20·(-405)
Δ = 32400
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{32400}=180$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-180}{2*20}=\frac{-180}{40} =-4+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+180}{2*20}=\frac{180}{40} =4+1/2 $
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