200=b^2+9b

Solution for 200=b^2+9b equation:

200=b^2+9b

We move all terms to the left:

200-(b^2+9b)=0

We get rid of parentheses

-b^2-9b+200=0

We add all the numbers together, and all the variables

-1b^2-9b+200=0

a = -1; b = -9; c = +200;
Δ = b2-4ac
Δ = -92-4·(-1)·200
Δ = 881
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{881}}{2*-1}=\frac{9-\sqrt{881}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{881}}{2*-1}=\frac{9+\sqrt{881}}{-2}
$