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-10n^2+n+3=0
a = -10; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-10)·3
Δ = 121
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{121}=11$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*-10}=\frac{-12}{-20} =3/5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*-10}=\frac{10}{-20} =-1/2 $
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