2x^2-8x+1=04

Solution for 2x^2-8x+1=04 equation:

2x^2-8x+1=04

We move all terms to the left:

2x^2-8x+1-(04)=0

We add all the numbers together, and all the variables

2x^2-8x-3=0

a = 2; b = -8; c = -3;
Δ = b2-4ac
Δ = -82-4·2·(-3)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{22}}{2*2}=\frac{8-2\sqrt{22}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{22}}{2*2}=\frac{8+2\sqrt{22}}{4}
$