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n^2+5n-16=0
a = 1; b = 5; c = -16;
Δ = b2-4ac
Δ = 52-4·1·(-16)
Δ = 89
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{89}}{2*1}=\frac{-5-\sqrt{89}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{89}}{2*1}=\frac{-5+\sqrt{89}}{2} $
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