28=x^2-4x

Solution for 28=x^2-4x equation:

28=x^2-4x

We move all terms to the left:

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

We get rid of parentheses

-x^2+4x+28=0

We add all the numbers together, and all the variables

-1x^2+4x+28=0

a = -1; b = 4; c = +28;
Δ = b2-4ac
Δ = 42-4·(-1)·28
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{2}}{2*-1}=\frac{-4-8\sqrt{2}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{2}}{2*-1}=\frac{-4+8\sqrt{2}}{-2}
$