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)#