n^2+n=200

Solution for n^2+n=200 equation:

n^2+n=200

We move all terms to the left:

n^2+n-(200)=0

a = 1; b = 1; c = -200;
Δ = b2-4ac
Δ = 12-4·1·(-200)
Δ = 801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{89}}{2*1}=\frac{-1-3\sqrt{89}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{89}}{2*1}=\frac{-1+3\sqrt{89}}{2}
$