Question 1189159
Solve the roots of the equation by completing the square. Leave all answers in exact values with the most simplified form:

3x^2 + 2x + 7/3 = 0
<pre>{{{matrix(1,3, 3x^2 + 2x + 7/3, "=", 0)}}}
{{{matrix(1,3, 3(x^2 + (2/3)x + 7/9), "=", 3(0))}}} -- Factoring out 3, to make the coefficient on the x<sup>2</sup>, 1
{{{matrix(1,3, x^2 + (2/3)x + 7/9, "=", 0)}}}
{{{matrix(1,3, x^2 + (2/3)x, "=", - 7/9)}}} ---------- Subtracting {{{7/9}}} from each side, to move constant to right of equals sign
{{{matrix(1,3, x^2 + (2/3)x + (1/3)^2, "=", - 7/9 + (1/3)^2)}}} ------ Taking {{{matrix(1,11, 1/2, of, b, "or", 1/2, of, "+", 2/3, "=", "+", 1/3)}}}, squaring it, and then ADDING the result to both sides of equation
{{{matrix(1,3, (x + 1/3)^2, "=", - 7/9 + 1/9)}}}
{{{matrix(2,3, (x + 1/3)^2, "=", - 6/9, (x + 1/3)^2, "=", - 2/3)}}}
{{{matrix(1,3, sqrt((x + 1/3)^2), "=", " " +-sqrt(- 2/3))}}} ------- Taking square root of both sides
{{{matrix(1,3, x + 1/3, "=", " " +-sqrt((2/3) * - 1))}}} 
{{{matrix(1,3, x + 1/3, "=", " " +-sqrt(2/3)i)}}} 

{{{highlight_green(system(matrix(1,3, x, "=", - 1/3 +-sqrt(2/3)i), highlight(or), matrix(1,7, x, "=", - 1/3 + sqrt(2/3)i, and, x, "=", - 1/3 - sqrt(2/3)i), highlight(or),
matrix(1,11, x, "=", - 1/3 + (sqrt(6)/3)i, and, x, "=", - 1/3 - (sqrt(6)/3)i, 
"------", Rationalizing, the, denominator)))}}} 

<font color = blue><font size = 4><b>Those are the EXACT solutions. Accept no other answers.</font></font></b></pre>