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2n^2+40n+182=0
a = 2; b = 40; c = +182;
Δ = b2-4ac
Δ = 402-4·2·182
Δ = 144
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{144}=12$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-12}{2*2}=\frac{-52}{4} =-13 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+12}{2*2}=\frac{-28}{4} =-7 $
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