<s>[INST] How to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$. I'm trying to figure out how to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$. 
Hints? 
 [/INST] HINT:
Complete the square, which can be done by noting that:
$$6x^2 - 20x + 23 = 6(x^2 - \frac{10}{3}x + \frac{25}{9}) +  \frac{19}{3}$$
Then multiply $(x^2 - \frac{10}{3}x + \frac{25}{9})$ by $9$ and divide $6$ by $9$.
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