Question 995565
Good Morning. I am trying to figure out how to complete the square for this problem but I am having no luck My problem is:
3x^2-13x+4=0. I know I have to divide each term by 3 which would give me:

x^2-13/3x+4/3=0. Then I subtract 4/3 on both sides to get x^2-13/3x=-4/3. I multiply -13/3xand 1/2 to get -13/6x then I square it to get 169/36 which I add to both sides of the equation. 
x^2-13/3x+169/36=-4/3+169/36. Here is where I'm having a hard time and are looking for your help.
<pre>{{{x^2 - (13/3)x + (- 13/6)^2 = - 4/3 + (- 13/6)^2}}} ------ This is where you are, Great job in getting to this point!
{{{(x - 13/6)^2 = - 4/3 + 169/36}}} ------ Factoring left-side of equation
{{{(x - 13/6)^2 = - 48/36 + 169/36}}} ----- Changing fraction on right to a common denominator, 36
{{{(x - 13/6)^2 = 121/36}}} ----------- Summing right side of equation
{{{sqrt((x - 13/6)^2) = " "+-sqrt(121/36)}}} ----- Taking the square root of both sides
{{{x - 13/6 = " "+- 11/6}}}
{{{x = 13/6 +- 11/6}}} ------- Adding {{{13/6}}} to both sides
{{{x = 13/6 + 11/6}}}		OR	   {{{x = 13/6 - 11/6}}}
{{{x = 24/6}}}         		OR	   {{{x = 2/6}}}
{{{highlight_green(system(x = 4_OR,x = 1/3))}}}