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3n^2+7n-520=0
a = 3; b = 7; c = -520;
Δ = b2-4ac
Δ = 72-4·3·(-520)
Δ = 6289
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{-(7)-\sqrt{6289}}{2*3}=\frac{-7-\sqrt{6289}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{6289}}{2*3}=\frac{-7+\sqrt{6289}}{6} $
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