x(22+x)=196

Solution for x(22+x)=196 equation:

x(22+x)=196

We move all terms to the left:

x(22+x)-(196)=0

We add all the numbers together, and all the variables

x(x+22)-196=0

We multiply parentheses

x^2+22x-196=0

a = 1; b = 22; c = -196;
Δ = b2-4ac
Δ = 222-4·1·(-196)
Δ = 1268
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:

$\sqrt{\Delta}=\sqrt{1268}=\sqrt{4*317}=\sqrt{4}*\sqrt{317}=2\sqrt{317}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{317}}{2*1}=\frac{-22-2\sqrt{317}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{317}}{2*1}=\frac{-22+2\sqrt{317}}{2}
$