2x^2-2x+4=12

Solution for 2x^2-2x+4=12 equation:

2x^2-2x+4=12

We move all terms to the left:

2x^2-2x+4-(12)=0

We add all the numbers together, and all the variables

2x^2-2x-8=0

a = 2; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·2·(-8)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{17}}{2*2}=\frac{2-2\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{17}}{2*2}=\frac{2+2\sqrt{17}}{4}
$