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3n^2+13n-10=0
a = 3; b = 13; c = -10;
Δ = b2-4ac
Δ = 132-4·3·(-10)
Δ = 289
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{289}=17$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*3}=\frac{-30}{6} =-5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*3}=\frac{4}{6} =2/3 $
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