Question 423835
Both equations are quadratic equations because of the squared terms. With quadratic  equationss you start by making one side zero and then factoring (or use the Quadratic Formula).
{{{2x^2=12}}}
{{{2x^2-12 = 0}}}
{{{2(x^2-6) = 0}}}
Normally one would say that {{{x^2-6}}} will not factor further. But let's go ahead and treat it as a difference of squares and factor it with the {{{a^2-b^2 = (a+b)(a-b)}}} pattern. But, you may ask, how is 6 a perfect square? Answer: It is not a square of an integer. But it is the square of something. It is the square of {{{sqrt(6)}}}! So we can factor {{{x^2-6}}} with the "b" being {{{sqrt(6)}}}:
{{{2(x+sqrt(6))(x-sqrt(6)) = 0}}}
From the Zero Product Property we know that one of these factors must be zero. The 2 is a 2, of course, and cannot be zero. But the other two factors can:
{{{x+sqrt(6) = 0}}} or {{{x-sqrt(6) = 0}}}
Solving these we get:
{{{x = -sqrt(6)}}} or {{{x = sqrt(6)}}}<br>
{{{147x^2=27}}}
{{{147x^2-27=0}}}
{{{3(49x^2-9)=0}}}
This time we have an obvious difference of squares:
3(7x+3)(7x-3)=0
7x+3 = 0 or 7x-3 = 0
x = -3/7 or x = 3/7