Question 83494
Start by putting your equation into the "standard form" for quadratic equations: {{{ax^2+bx+c = 0}}}
{{{6x = 3x^2-2}}} Subtract 6x from both sides.
{{{3x^2-6x-2 = 0}}} Now you need to make the coefficient of the {{{x^2}}} term equal to 1. You do this here by dividing through by 3.
{{{3x^2/3-6x/3-2/3 = 0/3}}} Simplify this.
{{{x^2-2x-2/3 = 0}}} Next, you'll add {{{2/3}}} to both sides.
{{{x^2-2x = 2/3}}} Now you'll complete the square by adding the square of half the x-coefficient {{{(-2/2)^2 = 1}}} to both sides.
{{{x^2-2x+1 = (2/3)+1}}} Simplify this.
{{{x^2-2x+1 = 5/3}}} Now factor the left side.
{{{(x-1)(x-1) = 5/3}}} Simplify this.
{{{(x-1)^2 = 5/3}}} Notice that you now have a "square" on the left side. Take the square root of both sides.
{{{x-1 = sqrt(5/3)}}} or {{{x-1 = -sqrt(5/3)}}} Add 1 to both sides of each equation.
{{{x = 1+sqrt(5/3)}}} or {{{x = 1-sqrt(5/3)}}} These are the two solutions.