x^2+22x=200

Solution for x^2+22x=200 equation:

x^2+22x=200

We move all terms to the left:

x^2+22x-(200)=0

a = 1; b = 22; c = -200;
Δ = b2-4ac
Δ = 222-4·1·(-200)
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{321}}{2*1}=\frac{-22-2\sqrt{321}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{321}}{2*1}=\frac{-22+2\sqrt{321}}{2}
$