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3n^2-53n+132=0
a = 3; b = -53; c = +132;
Δ = b2-4ac
Δ = -532-4·3·132
Δ = 1225
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{1225}=35$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-35}{2*3}=\frac{18}{6} =3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+35}{2*3}=\frac{88}{6} =14+2/3 $
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