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3n^2+n-690=0
a = 3; b = 1; c = -690;
Δ = b2-4ac
Δ = 12-4·3·(-690)
Δ = 8281
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{8281}=91$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-91}{2*3}=\frac{-92}{6} =-15+1/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+91}{2*3}=\frac{90}{6} =15 $
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