Completing the square problem The problem i am supposed to solve for $x$ by completing the square: 
$3x^2+9x+5 = 0$
step 1. move constant to right: $3x^2+9x\quad\quad+5 = 0$
step 2. divide by $3$: $x^2+3\quad\quad+\frac{5}{3}$
step 3. $(\frac{1}{2}b)^2$:  $(\frac{1}{2}\cdot 3)^2 = \frac{9}{4}$
step 4. add and subtract result from step 3 into equation: $(x^2+3+\frac{9}{4}) -\frac{9}{4} + \frac{5}{3} = 0$
step 5. simplify: $(x+\frac{3}{2})^2 -\frac{7}{12} = 0$
step 6. subtract constant on both sides: $(x+\frac{3}{2})^2 = \frac{7}{12}$
step 7. solve for $x$: $x = -\frac{3}{2} + \sqrt{\frac{7}{12}}$ or $-\frac{3}{2} - \sqrt{\frac{7}{12}}$
Wolfram seems to be telling me my answer is wrong.
the correct answer listed is: $x = \frac{-9-\sqrt{21}}{6}$ or $x = \frac{\sqrt{21}-9}{6}$
If anyone could let me know where my error is I would greatly appreciate it!
 A: As Theo Buehler said in a comment, your answer and technique are correct, just yielding a different form of the same answer (rationalize the denominator in your answer, then combine the fractions by finding a common denominator).
As an aside, while I'm sure that you're applying the technique as you were taught (those steps are fairly common in first-year algebra courses), I prefer a slightly different process for completing the square.  The process, described below, is a bit more compatible with uses of completing the square that show up in later courses.  Using a quadratic expression (not the whole equation):


*

*$3x^2+9x+5$

*Factor out the leading coefficient from the variable terms: $3(x^2+3x)+5$

*Determine the desired perfect-square constant term: $(\frac{1}{2}\cdot 3)^2=\frac{9}{4}$

*Add and subtract the desired constant term inside the parentheses: $3(x^2+3x+\frac{9}{4}-\frac{9}{4})+5$

*Factor the perfect square and distribute the leading coefficient: $3(x+\frac{3}{2})^2-3\cdot\frac{9}{4}+5$

*Combine the constant terms: $3(x+\frac{3}{2})^2-\frac{7}{4}$

