If a quadratic equation has solutions #x=a# and #x=b#, then #x-a=0# and #x-b=0#. Furthermore, the quadratic can be written as #y=c(x-a)(x-b)#, where #c# is some constant. The reasoning is that if you set #y# equal to #0#, you get:

#c(x-a)(x-b)=0#

Which is the same as:

#(x-a)(x-b)=0#

And so the solutions are #x=a# and #x=b# - which is exactly what we started with.

Alright, enough theory - let's get on with it! We are told that the #x#-intercepts are #-3# and #2#, and since #x#-intercepts are the same thing as zeros, #x=-3# and #x=2# are solutions. Following the process from above, we can write the quadratic as:

#y=c(x+3)(x-2)#

To solve for #c#, we use the other piece of info we were given: the point #(3,6)#:

#y=c(x+3)(x-2)#

#->6=c(3+3)(3-2)#

#->6=c(6)(1)#

#->6=6c->c=1#

So the equation of the quadratic is:

#y=1(x+3)(x-2)#

#->y=(x+3)(x-2)=x^2+3x-2x-6=x^2+x-6#