[InSovietRussiaORGASMGotU]

The table you're describing seems to be a table of values, although I do it in a slightly simpler way. As you know, to graph you need values in the form of (x,y). The table of values generally has x on the left and in the other column, it has y = [equation]. This allows you to get the values as (x,y).

y = x^2 - 4 1
y + x^2 = 4 2

It's much easier to make the equation have only 1 variable. I labelled them 1 and 2 just so it's simpler for me to write.

Sub 1 into 2:
x^2-4 + x^2 = 4
2x^2 = 8
x^2 = 4
x = +/-2

Since 1 has x^2, we'll get the same y-value regardless if x = -2 or x = 2:
Sub x = 2 into 1:
y = 2^2-4
y = 0

Therefore, solution of (2,0) and (-2,0).

For the table format, choose whatever x-values you want and plug them into 1.

y = x^2 - 5x + 8 1
y = -x^2 + 7x - 10 2

In 1 and 2, we see both have y = blah blah and y = blah blah. So, make it y = y:
x^2-5x+8 = -x^2+7x-10
2x^2-12x + 18 = 0
We see that we can factor out a 2 to make it simpler:
2(x^2-6x+9) = 0
x^2-6x+9 = 0
We can factor this out through common factoring, Quadratic Equation, graphing calculator, whatever one you really want. The basic idea though is choose 2 numbers, and see if they add to the middle one (-6) and if they multiply to give the third one (9).
We can choose either 1x9 or 3x3 (or decimals or whatever but common sense tells us we won't need that).
1 and 9 cannot give us a -6.
-3+(-3) can give us -6.
-3x-3 = 9
So,
x^2-6x+9 = 0 factors to (x-3)(x-3) = 0 or simply (x-3)^2 = 0.
To get any y-values, put the x-values into 1 or 2. I believe they'll give you the same answer, so pick which one you want.

For both of these questions, you can graph the equations by hand, although you'll need to do more calculations, which I assume you're not doing right now.