There are a few ways to do this. There is an easy way for this kind of question with "the brackets", but you might get confused since it isn't the usual method that you would use to solve quadratics.
Ok so I'll try and explain it to the best of my ability the simple way first:
You have y=(x+1)^2-2
Now, you know when a number is squared a^2 the value will always be zero or positive, because whether "a" is positive or negative, square it and it becomes positive.
This means that the (x+1)^2 will always be 0 or more. When is it the smallest though? Well that happens when the part inside the brackets is equal to zero. So x+1=0 and thus x=-1
Now since every other value of x will make the part inside the bracket not zero, when you square this non-zero number it will be positive. This means that the y value will grow larger as x moves away from -1. So that means the parabola has a minimum turning point when the y value is the smallest. And this happens when x=-1. So what is the smallest y value? Well when x=-1 the bracket part is 0 and all that is left is y=0-2=-2
So now you have the minimum turning point (-1,-2)
If you're still reading up to this point, then you have determination, and I'll keep going
As for the way the teacher was asking you to do it, which I think is not the way you were trying to solve the quadratic.
Your teacher said "factorize by sight in pairs"
y=(x+1)^2-2
Notice how this is equivalent to y=a^2-b^2
And if you factorize by difference of 2 squares: y=(a+b)(a-b)
We just need to find what a and b are.
Well if (x+1)^2-2\equiv a^2-b^2
Then (x+1)^2=a^2
So x+1=a
And 2=b^2
So b=\sqrt{2}
(don't worry about the \pm here though, as you'll notice later)
So substituting the a and b values into the general difference of 2 squares formula thingy:
y=(x+1+\sqrt{2})(x+1-\sqrt{2})
So you've now factorized the quadratic (don't worry, it will become way easier than it looks here right now)
So find the roots of the quadratic (or the zeros, simply where it cuts the x-axis).
This is when y=0, so 0=(x+1+\sqrt{2})(x+1-\sqrt{2})
Now solve for x like you would've done with all the other quadratics. Don't be afraid of the irrationals, they work just like any other number.
And notice that if you also take the negative of (x+1) and \sqrt{2} because of the \pm that we ignored earlier, it will give you the 2 same answers.
Now that you have the roots, find the midpoint to give you the x-value where the turning point is and then substitute that x value into the quadratic to give you the y value.
And finally, just to clear you up on the way you were trying to do it.
The formula y=ax^2+bx+c needs to be followed in the exact way it is shown. What I mean is, all the values with x^2 need to be together, all the values with x need to be together, and all the constants need to be together.
e.g. y=2x^2+3x+x^2-5-\sqrt{5}x+1+\sqrt{2}
You need to collect all co-efficients (numbers in front of) the x^2s, xs and all the numbers. (EVEN if they cannot be expressed as 1 number).
So, putting all equal variables (x is the variable) together:
y=2x^2+x^2+3x-\sqrt{5}x+1-5+\sqrt{2}
And obviously things like 2x^2+x^2=3x^2 but it would be best to factorize x^2 out first just to get the idea:
y=(2+1)x^2+(3-\sqrt{5})x+(1-5+\sqrt{2})
and adding terms as much as possible:
y=3x^2+(3-\sqrt{5})x-4+\sqrt{2}
Now look at the formula again: y=ax^2+bx+c
The values a, b and c must be equivalent so:
a=3 b=3-\sqrt{5} c=-4+\sqrt{2}
This is how you will now substitute into the quadratic formula to find the roots of the quadratic. When you try it for your question, you first need to expand and then collect all the terms and put them into the same form as I've shown.
Ok I think I've written enough... *walks away quietly*
