n^2+n-1332=0

Solution for n^2+n-1332=0 equation:

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
$