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3n^2-53n+232=0
a = 3; b = -53; c = +232;
Δ = b2-4ac
Δ = -532-4·3·232
Δ = 25
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{25}=5$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-5}{2*3}=\frac{48}{6} =8 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+5}{2*3}=\frac{58}{6} =9+2/3 $
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