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3n^2+3n-200=0
a = 3; b = 3; c = -200;
Δ = b2-4ac
Δ = 32-4·3·(-200)
Δ = 2409
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{-(3)-\sqrt{2409}}{2*3}=\frac{-3-\sqrt{2409}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{2409}}{2*3}=\frac{-3+\sqrt{2409}}{6} $
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