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n2+n-1332=0
We add all the numbers together, and all the variables
n^2+n-1332=0
a = 1; b = 1; c = -1332;
Δ = b2-4ac
Δ = 12-4·1·(-1332)
Δ = 5329
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{5329}=73$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-73}{2*1}=\frac{-74}{2} =-37 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+73}{2*1}=\frac{72}{2} =36 $
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