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3n^2-25n+40=0
a = 3; b = -25; c = +40;
Δ = b2-4ac
Δ = -252-4·3·40
Δ = 145
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{145}}{2*3}=\frac{25-\sqrt{145}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{145}}{2*3}=\frac{25+\sqrt{145}}{6} $
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