a^2-2a+1=10

Solution for a^2-2a+1=10 equation:

a^2-2a+1=10

We move all terms to the left:

a^2-2a+1-(10)=0

We add all the numbers together, and all the variables

a^2-2a-9=0

a = 1; b = -2; c = -9;
Δ = b2-4ac
Δ = -22-4·1·(-9)
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{10}}{2*1}=\frac{2-2\sqrt{10}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{10}}{2*1}=\frac{2+2\sqrt{10}}{2}
$