we have
[tex]y=x^{2} +6x+10[/tex]
we know that
the equation in vertex form is equal to
[tex]y=(x-h)^{2} +k\\[/tex]
where
[tex](h,k)[/tex] is the vertex
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]y-10=x^{2} +6x[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]y-10+9=x^{2} +6x+9[/tex]
[tex]y-1=x^{2} +6x+9[/tex]
Rewrite as perfect squares
[tex]y-1=(x+3)^{2}[/tex]
[tex]y=(x+3)^{2}+1[/tex]
[tex](h,k)=(-3,1)[/tex]
therefore
the answer is
the equation in vertex form is equal to
[tex]y=(x+3)^{2}+1[/tex]
