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4n^2-13n+3=0
a = 4; b = -13; c = +3;
Δ = b2-4ac
Δ = -132-4·4·3
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*4}=\frac{2}{8} =1/4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*4}=\frac{24}{8} =3 $
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