4=x^2+4x

Solution for 4=x^2+4x equation:

4=x^2+4x

We move all terms to the left:

4-(x^2+4x)=0

We get rid of parentheses

-x^2-4x+4=0

We add all the numbers together, and all the variables

-1x^2-4x+4=0

a = -1; b = -4; c = +4;
Δ = b2-4ac
Δ = -42-4·(-1)·4
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{2}}{2*-1}=\frac{4-4\sqrt{2}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{2}}{2*-1}=\frac{4+4\sqrt{2}}{-2}
$