20x+x^2=200

Solution for 20x+x^2=200 equation:

20x+x^2=200

We move all terms to the left:

20x+x^2-(200)=0

a = 1; b = 20; c = -200;
Δ = b2-4ac
Δ = 202-4·1·(-200)
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{3}}{2*1}=\frac{-20-20\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{3}}{2*1}=\frac{-20+20\sqrt{3}}{2}
$