x^2-4x+8=8+4x-x^2

Solution for x^2-4x+8=8+4x-x^2 equation:

x^2-4x+8=8+4x-x^2

We move all terms to the left:

x^2-4x+8-(8+4x-x^2)=0

We get rid of parentheses

x^2+x^2-4x-4x-8+8=0

We add all the numbers together, and all the variables

2x^2-8x=0

a = 2; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·2·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*2}=\frac{16}{4}
=4
$