n^2+n-1122=0

Solution for n^2+n-1122=0 equation:

n^2+n-1122=0

a = 1; b = 1; c = -1122;
Δ = b2-4ac
Δ = 12-4·1·(-1122)
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-67}{2*1}=\frac{-68}{2}
=-34
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+67}{2*1}=\frac{66}{2}
=33
$