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3n^2+14n-369=0
a = 3; b = 14; c = -369;
Δ = b2-4ac
Δ = 142-4·3·(-369)
Δ = 4624
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{4624}=68$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-68}{2*3}=\frac{-82}{6} =-13+2/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+68}{2*3}=\frac{54}{6} =9 $
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