Question 166950
<font size = 7 color = "red"><b>Edwin's solution:</b></font>
<pre><font size = 4 color = "indigo"><b>

{{{2x^2-3x+1=0}}}

Isolate the variable terms by
adding -1 to both sides:

{{{2x^2-3x=-1}}}

Divide thru by 2, the coefficient of the 
term in {{{x^2}}}

{{{(2/2)x^2-(3/2)x=-1/2}}}

{{{x^2-(3/2)x=-1/2}}}

To the side, multiply the coefficient of x by {{{1/2}}}.
{{{-3/2)(1/2)=-3/4}}}

Now square {{{-3/4}}}.  {{{(-3/4)^2=9/16}}}

Add {{{9/16}}} to both sides of

{{{x^2-(3/2)x+9/16=-1/2+9/16}}}

Combine the right side: {{{-1/2+9/16=-8/16+9/16=1/16}}}

{{{x^2-(3/2)x+9/16=1/16}}}

Factor the left side:

{{{(x-3/4)(x-3/4)=1/16}}}

Write the left side as the square of
a binomial:

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

Take the square root of both sides,
remembering the ± on the right:

{{{sqrt((x-3/4)^2) = "" +-sqrt(1/16)}}}

{{{x-3/4}}}=±{{{1/4}}}

{{{x=3/4+-1/4}}}

Using the +

{{{x=3/4+1/4}}}
{{{x=4/4}}}
{{{x=1}}}

Using the -

{{{x=3/4-1/4}}}
{{{x=2/4}}}
{{{x=1/2}}}

The solutions are {{{1}}} and {{{1/2}}}

Edwin</pre>