x^2+2x+1=200

Solution for x^2+2x+1=200 equation:

x^2+2x+1=200

We move all terms to the left:

x^2+2x+1-(200)=0

We add all the numbers together, and all the variables

x^2+2x-199=0

a = 1; b = 2; c = -199;
Δ = b2-4ac
Δ = 22-4·1·(-199)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-20\sqrt{2}}{2*1}=\frac{-2-20\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+20\sqrt{2}}{2*1}=\frac{-2+20\sqrt{2}}{2}
$