We start with #-2x^2+3x-6#.
The way I would solve this is by completing the square. The first step for that is to make the coefficient of #x^2# 1. We do that by factoring out a #-2#. The equation now looks like this:
#-2(x^2-3/2x+3)#.
From here, we need to find a term which will make the equation factorable. We do that by taking the middle factor, #-3/2#, and dividing it by #2#, making it #-3/4#. Then we square this, changing it to #9/16#.
Now that we found the number which will make the#x^2-3/2#part of the equation factorable, what do we do with it? I'll tell you what we do with it; we plug it in . But, we can't just put a random number into the equation. I'll show you how we resolve this in a minute.
First we rewrite the equation as #-2(x^2-3/2color(red)(+9/16)color(red)(-9/16)+3 )#. NOTE we resolved the issue of sticking in a number by subtracting it so that it actually has no effect on the value of the equation.
Anyway, we can now condense #-2(x^2-3/2+9/16-9/16+3 )# into #-2((x-3/4)^2+39/16)#.
We are very nearly done, except that we can stil simplify further by multiplying #-2# to the #39/16#, making it #-39/8#.
The final answer is #-2(x-3/4)^2-39/8=y#
