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3n^2-5n-10=0
a = 3; b = -5; c = -10;
Δ = b2-4ac
Δ = -52-4·3·(-10)
Δ = 145
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{145}}{2*3}=\frac{5-\sqrt{145}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{145}}{2*3}=\frac{5+\sqrt{145}}{6} $
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