Answer:
[tex]\boxed{\bold{y=(x+3)^2-14}}[/tex]
Step-by-step explanation:
y= x²+6x-5
completing the square: a²+2ab+b² = (a+b)² where a²=x² (so a=x) and 2ab=6x (so b would be 3):
[tex]y=x^2+6x-5\\\\y=\underline{x^2+2\cdot x\cdot3+3^2}-3^2-5\\\\y=(x+3)^2-9-5\\\\\underline{y=(x+3)^2-14}\\\\\\vertex:\ \ (-3,\,-14)[/tex]
