x2+4x=196

Solution for x2+4x=196 equation:

x2+4x=196

We move all terms to the left:

x2+4x-(196)=0

We add all the numbers together, and all the variables

x^2+4x-196=0

a = 1; b = 4; c = -196;
Δ = b2-4ac
Δ = 42-4·1·(-196)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20\sqrt{2}}{2*1}=\frac{-4-20\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20\sqrt{2}}{2*1}=\frac{-4+20\sqrt{2}}{2}
$