(0.5x^2)-2x+1=0

Solution for (0.5x^2)-2x+1=0 equation:

(0.5x^2)-2x+1=0

We add all the numbers together, and all the variables

-2x+(0.5x^2)+1=0

We get rid of parentheses

0.5x^2-2x+1=0

a = 0.5; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·0.5·1
Δ = 2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-\sqrt{2}}{2*0.5}=\frac{2-\sqrt{2}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{2}}{2*0.5}=\frac{2+\sqrt{2}}{1}
$