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3n^2-39n+90=0
a = 3; b = -39; c = +90;
Δ = b2-4ac
Δ = -392-4·3·90
Δ = 441
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{441}=21$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-21}{2*3}=\frac{18}{6} =3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+21}{2*3}=\frac{60}{6} =10 $
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