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5n^2+19n-72=0
a = 5; b = 19; c = -72;
Δ = b2-4ac
Δ = 192-4·5·(-72)
Δ = 1801
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{-(19)-\sqrt{1801}}{2*5}=\frac{-19-\sqrt{1801}}{10} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{1801}}{2*5}=\frac{-19+\sqrt{1801}}{10} $
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