2x^2-4x+1=25

Solution for 2x^2-4x+1=25 equation:

2x^2-4x+1=25

We move all terms to the left:

2x^2-4x+1-(25)=0

We add all the numbers together, and all the variables

2x^2-4x-24=0

a = 2; b = -4; c = -24;
Δ = b2-4ac
Δ = -42-4·2·(-24)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{13}}{2*2}=\frac{4-4\sqrt{13}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{13}}{2*2}=\frac{4+4\sqrt{13}}{4}
$