Algebra 2

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

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asked by Keith
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