What is y= 6x^2 + 12x -10 in vertex form?

Vertex for  is a(x - b)^2 + c

first divide 6x^2 + 12x by 6  so we have

6(x^2 + 2x) - 10

now complete the square  on the expression in the brackets:-
= 6[ (x + 1)^2 - 1] -10
=6(x + 1)^2 - 6 -10
= 6( x+ 1)^2 - 16

Basically complete the square
so
conver tto y=a(x-h)²+k
steps:
group x terms
factor out leading coefient
take 1/2 of ilnear coefient and square it, add negative and positive inside
factor perfect square
expand/distribute

y=6x²+12x-10

group x
y=(6x²+12x)-10

factor out leading coefient
y=6(x²+2x)-10

take 1/2 of linear coefient and square it
2/2=1, 1²=1
add negativ and positivve inside
y=6(x²+2x+1-1)-10

factor perfect square
y=6((x+1)²-1)-10

distribute
y=6(x+1)²-6-10
y=6(x+1)²-16

The answer is a hope i