32x^2-64x+32=1

Solution for 32x^2-64x+32=1 equation:

32x^2-64x+32=1

We move all terms to the left:

32x^2-64x+32-(1)=0

We add all the numbers together, and all the variables

32x^2-64x+31=0

a = 32; b = -64; c = +31;
Δ = b2-4ac
Δ = -642-4·32·31
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-8\sqrt{2}}{2*32}=\frac{64-8\sqrt{2}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+8\sqrt{2}}{2*32}=\frac{64+8\sqrt{2}}{64}
$