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2n^2+25n-13=0
a = 2; b = 25; c = -13;
Δ = b2-4ac
Δ = 252-4·2·(-13)
Δ = 729
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{729}=27$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-27}{2*2}=\frac{-52}{4} =-13 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+27}{2*2}=\frac{2}{4} =1/2 $
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