2x^2-4x+4=56

Solution for 2x^2-4x+4=56 equation:

2x^2-4x+4=56

We move all terms to the left:

2x^2-4x+4-(56)=0

We add all the numbers together, and all the variables

2x^2-4x-52=0

a = 2; b = -4; c = -52;
Δ = b2-4ac
Δ = -42-4·2·(-52)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12\sqrt{3}}{2*2}=\frac{4-12\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12\sqrt{3}}{2*2}=\frac{4+12\sqrt{3}}{4}
$