2n2+2n=1800

Solution for 2n2+2n=1800 equation:

2n^2+2n=1800

We move all terms to the left:

2n^2+2n-(1800)=0

a = 2; b = 2; c = -1800;
Δ = b2-4ac
Δ = 22-4·2·(-1800)
Δ = 14404
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:

$\sqrt{\Delta}=\sqrt{14404}=\sqrt{4*3601}=\sqrt{4}*\sqrt{3601}=2\sqrt{3601}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3601}}{2*2}=\frac{-2-2\sqrt{3601}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3601}}{2*2}=\frac{-2+2\sqrt{3601}}{4}
$