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5n^2+11n-260=0
a = 5; b = 11; c = -260;
Δ = b2-4ac
Δ = 112-4·5·(-260)
Δ = 5321
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{-(11)-\sqrt{5321}}{2*5}=\frac{-11-\sqrt{5321}}{10} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{5321}}{2*5}=\frac{-11+\sqrt{5321}}{10} $
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