199.5=0.5n^2+n

Solution for 199.5=0.5n^2+n equation:

199.5=0.5n^2+n

We move all terms to the left:

199.5-(0.5n^2+n)=0

We get rid of parentheses

-0.5n^2-n+199.5=0

We add all the numbers together, and all the variables

-0.5n^2-1n+199.5=0

a = -0.5; b = -1; c = +199.5;
Δ = b2-4ac
Δ = -12-4·(-0.5)·199.5
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-20}{2*-0.5}=\frac{-19}{-1}
=+19
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+20}{2*-0.5}=\frac{21}{-1}
=-21
$