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3n^2-29n+48=0
a = 3; b = -29; c = +48;
Δ = b2-4ac
Δ = -292-4·3·48
Δ = 265
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{-(-29)-\sqrt{265}}{2*3}=\frac{29-\sqrt{265}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{265}}{2*3}=\frac{29+\sqrt{265}}{6} $
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