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2n^2+2n-1300=0
a = 2; b = 2; c = -1300;
Δ = b2-4ac
Δ = 22-4·2·(-1300)
Δ = 10404
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{10404}=102$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-102}{2*2}=\frac{-104}{4} =-26 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+102}{2*2}=\frac{100}{4} =25 $
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