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n^2+16n+28=0
a = 1; b = 16; c = +28;
Δ = b2-4ac
Δ = 162-4·1·28
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-12}{2*1}=\frac{-28}{2} =-14 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+12}{2*1}=\frac{-4}{2} =-2 $
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