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3n^2-11n-36=0
a = 3; b = -11; c = -36;
Δ = b2-4ac
Δ = -112-4·3·(-36)
Δ = 553
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{-(-11)-\sqrt{553}}{2*3}=\frac{11-\sqrt{553}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{553}}{2*3}=\frac{11+\sqrt{553}}{6} $
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