Question 926519
The steps to completing the square are the same for each of these, but the first one contains one less step since the coefficient of {{{x^2}}} is 1.

For #1, the first step is to move the constant to the other side.

{{{x^2 - 6x = 10}}}

Now, we take the b term (-6 in this case), divide it by half and square it, and add it to both sides: (b/2)^2.

{{{x^2 - 6x + (-6/2)^2 = 10 + (-6/2)^2}}}

Now, this past step created what is called a perfect square trinomial, which allows us to factor the left hand side as {{{(x + (b/2))^2}}}, so we have:

{{{(x + (-6/2))^2 = 10 + 9}}}

Simplifying a bit we get

{{{(x - 3)^2 = 19}}}

Since we now have something squared equal to a constant we can take the square root of both sides to obtain:

{{{x - 3 = +- sqrt(19)}}}

Adding 3 to both sides we get our final answer:

{{{x = 3 +- sqrt(19)}}}

That's it!!

The only difference in #2 is that we must divide through by 3 in order for the coefficient of the {{{x^2}}} term to be 1.  So, step by step we have:

{{{3x^2 - 6x = 6}}}

{{{x^2 - 2x = 2}}}

{{{x^2 - 2x + (-2/2)^2 = 2 + (-2/2)^2}}}

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

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

{{{x - 1 = +- sqrt(3)}}}

{{{x = 1 +- sqrt(3)}}}

Hope that helps!!