200X+x^2=4400

Solution for 200X+x^2=4400 equation:

200X+X^2=4400

We move all terms to the left:

200X+X^2-(4400)=0

a = 1; b = 200; c = -4400;
Δ = b2-4ac
Δ = 2002-4·1·(-4400)
Δ = 57600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{57600}=240$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-240}{2*1}=\frac{-440}{2}
=-220
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+240}{2*1}=\frac{40}{2}
=20
$