x²+5x+6 min or max

NOTE:You need the y= out the front or the question is meaningless!

CPhill has used the first bit of the quadratic formula to find the x value.

Another way is to factorise.

$$y=x^2+5x+6\\
y=(x+3)(x+2)$$

The roots are where y=0 and this is where x+3=0  or  x+2=0

Solving these gives you x=-3 and x=-2

The axis of symmetry will give you the x value of the vertex.

$$x=\frac{-3+-2}{2}=\frac{-5}{2}$$

Now the rest is exactly the same as CPhill's method.

I assune you want to find the vertex for this and also want to know if this is a minimum or maximum.

The x coordinate of the vertex is given by -b/(2a), where b = 5 and a =1.

So we have -5/(2(1)) = -5/2 = -2.5.

And "plugging" this back into the function, we find that the y coordinate of the vertex =

(-2.5)^2 + 5(-2.5) + 6 = -.25

Now, if you haven't had calculus yet, just trust me that something called the "second derivative" can be used to determine if the vertex is a minimum or a maximum.

If the second derivative > 0, we have a minimum.....and in this case, the second derivative = 2......so (-2.5, -.25) is a minimum......but....if you don't believe me look at the graph of the function  below......

The vertex is at (-2.5, -.25) and it is indeed a minimum .... just as we expected !!