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!