b^2+100b=200000

Solution for b^2+100b=200000 equation:

b^2+100b=200000

We move all terms to the left:

b^2+100b-(200000)=0

a = 1; b = 100; c = -200000;
Δ = b2-4ac
Δ = 1002-4·1·(-200000)
Δ = 810000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{810000}=900$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-900}{2*1}=\frac{-1000}{2}
=-500
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+900}{2*1}=\frac{800}{2}
=400
$