Given that $m$ is a real number not less than $-1$ The question is:
Given that $m$ is a real number not less than $-1$, such that the equation in $x$ is $x^2+2(m-2)x+m^2-3m+3=0$ has two distinct roots $r$ and $s$. If $r^2+s^2=6$, find the value of $m$.  
Here's what I've tried:. 
Using Vieta's formulas: 
$rs = m^2-3m+3$
$r+s = -2m+4$   
Then I set the equation as such; $(r+s)^2 - 2rs = 6$. I got $2m^2-10m+4=0$. Using the quadratic equation, I got $\frac {5+\sqrt {17}}2$ and $\frac {5-\sqrt {17}}2$. Is there anything wrong with my solution?

It is correct:
$$(r+s)^2 - 2rs = 6 \iff \\
(-2(m-2))^2-2(m^2-3m+3)=6 \iff \\
(4m^2-16m+16)-2m^2+6m-6=6 \iff \\
2m^2-10m+4=0.$$
And the roots are:
$$m_{1,2}=\frac{5\pm \sqrt{17}}{2}.$$
