How do you solve the quadratic equation by completing the square: `2x^2 - 3x + 1 = 0`?

The first thing you need to do is get your quadratic to the form

`x^2 + b/ax = -c/a`

To do that, add `-1` to both sides of the equation and divide every thing by 2, the coefficient of `x^2`.

`2x^2 - 3x + color(red)(cancel(color(black)(1))) - color(red)(cancel(color(black)(1))) = -1`

`(color(red)(cancel(color(black)(2))) x^2)/color(red)(cancel(color(black)(2))) -3/2x = -1/2`

`x^2 - 3/2x = -1/2`

The idea behing using the completing the square technique is that you need to write the left side of the equation as a square of a binomial by adding a term to both sides of the equation.

This term can be found by dividing the coefficient of the `x`-term by 2, then squaring it. In your case, you have

`(-3/2)* 1/2 = -3/4`, then

`(-3/4)^2 = 9/16`

Your quadratic now becomes

`x^2 - 3/2x + 9/16 = -1/2 + 9/16`

The left side of the equation can now be written as

`x^2 - 3/2x + 9/16 = x^2 - 2 * 3/4x + (-3/4)^2 = (x-3/4)^2`

This means that you have

`(x-3/4)^2 = 1/16`

Take the square root of boths sides to get

`sqrt((x-3/4)^2) = sqrt(1/16)`

`x-3/4 = +- 1/4 => x_(1,2) = 3/4 +- 1/4`

The two solutions of the quadratic will thus be

`x_1 = 3/4 - 1/4 = color(green)(1/2)` and `x_2 = 3/4 + 1/4 = color(green)(1)`