2x^2-4x+4=58

Solution for 2x^2-4x+4=58 equation:

2x^2-4x+4=58

We move all terms to the left:

2x^2-4x+4-(58)=0

We add all the numbers together, and all the variables

2x^2-4x-54=0

a = 2; b = -4; c = -54;
Δ = b2-4ac
Δ = -42-4·2·(-54)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{7}}{2*2}=\frac{4-8\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{7}}{2*2}=\frac{4+8\sqrt{7}}{4}
$