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2n^2=200
We move all terms to the left:
2n^2-(200)=0
a = 2; b = 0; c = -200;
Δ = b2-4ac
Δ = 02-4·2·(-200)
Δ = 1600
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{1600}=40$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*2}=\frac{-40}{4} =-10 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*2}=\frac{40}{4} =10 $
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