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10n^2+3n-1=0
a = 10; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·10·(-1)
Δ = 49
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{49}=7$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*10}=\frac{-10}{20} =-1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*10}=\frac{4}{20} =1/5 $
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