Question 35605
{{{x^2}}}- 3x + 2 = 0
1)Solve by factoring. 
First fiqure out the factors of 2; 2,1 
now added together they give us 3, so we have;
(x-2)(x-1)

2)Solve by completing the square.
We start by moving 2 to the right side;
{{{x^2}}}-3x=-2
The coefficient of x=-3; we divide -3 by 2 and square the result;
({{{-3/2}}})^2={{{9/4}}}
Add {{{9/4}}} to both sides;
{{{x^2}}}-3x+{{{9/4}}}=-2+{{{9/4}}}
simplified;
{{{(x-3/2)^2}}}={{{1/4}}}
square root both sides;
x-{{{3/2}}}=(+/-){{{sqrt(1/4))}}}
({{{sqrt(1/4)}}} = 1/2)
Add {{{3/2}}} to both sides;
x={{{3/2}}}(+/-){{{1/2}}}
x={{{3/2}}}+{{{1/2}}} or x={{{3/2}}}-{{{1/2}}}
x=2 or x=1

3)Solve by using the quadratic formula.
quadratic formula;
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=1; b=-3; c=2
{{{x=(-(-3)+- sqrt((-3)^2-4(1)(2)))/2(1)}}}
{{{x=(3+-sqrt(9-8))/2}}}
{{{x=(3+-sqrt(1))/2}}}
x={{{(3+-1)/2}}}
x={{{(3+1)/2}}} or x={{{(3-1)/2}}}
x=2 or x=1
Hope you understand everything
=)