I had to solve this quadratic equation using the complete the square method. Did I do it correctly?

The problem is 3x^2+18x+10= 0
(3x^2+18x+10)/3= 0/3
= x^2+6x+(10/3)=0
x^2+6x+(10/3)-(10/3)=0-(10/3)
=x^2+6x= (-10/3(
x^2+6x +(6/2)^2= -10/3+(6/2)^2
x^2+6x+9=(x+3)^2
(-10/3)+(9/1)
(-10/3)+(27/3)=17/3
x^2+6x+9 =17/3
(x+3^2)=17/3
√(x+3)^2=x+3
x+3=√17/3
x=-3+ √17/3
x=-3- √17/3
So my final answer is x= -3 ± √17/3

correct

Yes, you have correctly solved the quadratic equation using the complete the square method. The steps you followed are as follows:

1. Divide the entire equation by 3 to simplify: (3x^2 + 18x + 10)/3 = 0/3
This gives you: x^2 + 6x + (10/3) = 0

2. Move the constant term to the other side: x^2 + 6x + (10/3) - (10/3) = 0 - (10/3)
Simplifying this gives: x^2 + 6x = (-10/3)

3. Add the square of half the coefficient of x to both sides of the equation: x^2 + 6x + (6/2)^2 = (-10/3) + (6/2)^2
Simplifying this gives: x^2 + 6x + 9 = (-10/3) + (9/1)

4. Simplify the right side of the equation: (-10/3) + (9/1) = 17/3

5. Rewrite the left side of the equation as a perfect square: (x + 3)^2 = 17/3

6. Take the square root of both sides: √(x + 3)^2 = √(17/3)
This gives you: x + 3 = √(17/3)

7. Solve for x by subtracting 3 from both sides: x = -3 + √(17/3) and x = -3 - √(17/3)

So, your final answer is x = -3 ± √(17/3). Well done!