200=x^2+x

Solution for 200=x^2+x equation:

200=x^2+x

We move all terms to the left:

200-(x^2+x)=0

We get rid of parentheses

-x^2-x+200=0

We add all the numbers together, and all the variables

-1x^2-1x+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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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