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

Solution for 2x(x-4)+2=4x-2x^2 equation:

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

We move all terms to the left:

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2-12x+2=0

a = 4; b = -12; c = +2;
Δ = b2-4ac
Δ = -122-4·4·2
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{7}}{2*4}=\frac{12-4\sqrt{7}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{7}}{2*4}=\frac{12+4\sqrt{7}}{8}
$