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3n^2-53n+230=0
a = 3; b = -53; c = +230;
Δ = b2-4ac
Δ = -532-4·3·230
Δ = 49
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{49}=7$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-7}{2*3}=\frac{46}{6} =7+2/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+7}{2*3}=\frac{60}{6} =10 $
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