x^2+40x=200

Solution for x^2+40x=200 equation:

x^2+40x=200

We move all terms to the left:

x^2+40x-(200)=0

a = 1; b = 40; c = -200;
Δ = b2-4ac
Δ = 402-4·1·(-200)
Δ = 2400
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:

$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{6}}{2*1}=\frac{-40-20\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{6}}{2*1}=\frac{-40+20\sqrt{6}}{2}
$