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n^2+16n-57=0
a = 1; b = 16; c = -57;
Δ = b2-4ac
Δ = 162-4·1·(-57)
Δ = 484
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{484}=22$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-22}{2*1}=\frac{-38}{2} =-19 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+22}{2*1}=\frac{6}{2} =3 $
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