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2n^2+40n+38=0
a = 2; b = 40; c = +38;
Δ = b2-4ac
Δ = 402-4·2·38
Δ = 1296
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{1296}=36$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-36}{2*2}=\frac{-76}{4} =-19 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+36}{2*2}=\frac{-4}{4} =-1 $
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