Find the vertex of parabola f(x)=6x^2+12x+78?

The coordinates of the vertex of the parabola
are:

V(-b/2a ;
-delta/4a)

a,b,c are the coefficients of the
quadratic:

y = ax^2 + bx +
c

Comparing, we'll get the values of the
coefficients:

a = 6, b = 12 , c =
78

We'll determine the x coordinate of the
vertex:

xV = -b/2a

xV =
-12/12

xV = -1

We'll determine
the y coordinate of the vertex:

yV =
-delta/4a

delta = b^2 -
4ac

delta = 144 - 4*6*78

delta
= 144 - 1872

yV =
-(-1728)/24

yV =
72

The coordinates of the vertex of the
parabola are: V(-1, 72).