1200=2n+2n^2

Solution for 1200=2n+2n^2 equation:

1200=2n+2n^2

We move all terms to the left:

1200-(2n+2n^2)=0

We get rid of parentheses

-2n^2-2n+1200=0

a = -2; b = -2; c = +1200;
Δ = b2-4ac
Δ = -22-4·(-2)·1200
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-98}{2*-2}=\frac{-96}{-4}
=+24
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+98}{2*-2}=\frac{100}{-4}
=-25
$