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n^2+25n+156=0
a = 1; b = 25; c = +156;
Δ = b2-4ac
Δ = 252-4·1·156
Δ = 1
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{1}=1$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-1}{2*1}=\frac{-26}{2} =-13 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+1}{2*1}=\frac{-24}{2} =-12 $
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