I really can't figure out equations with a power  of 2 in it, please solve these and explain every step as if I was a baby.

Assuming that by x^2 (which I will write as $x^2$ from now on) you mean the operation of raising a number to the power of 2, here's a simple approach to problems like these:

Step 1: expand the products using the distributive property and paying attention to the signs.
$$ 3x -(x-6)(x-2) = 7x - x^2 \\ 3 x - (x (x - 2) - 6 (x - 2)) = 7x - x^2 \\ 3x - (xx - 2x - 6x+12) =7x - x^2 $$
Step 2: whenever you end up with some quantity multiplied by itself, like $xx$ above, you can rewrite that as the square of that quantity (in this case, $x^2$).
$$3x - (x^2 - 2x -6x+12) =7x - x^2$$
Step 3: remove any remaining parentheses by distributing the signs.
$$3x - x^2 + 2x + 6x - 12 = 7x - x^2$$
Step 4: bring everything over to one side by subtracting the value on the right side from both sides of the equation (remember that subtracting a negative value means adding its absolute value).
$$ 3x - x^2 + 2x +6x - 12 \ \underline{- 7x + x^2} = 7x - x^2 \ \underline{- 7x + x^2} \\ 3x - x^2 + 2x + 6x - 12 - 7x + x^2 = 0 $$
Step 5: add like terms together
$$ 4x - 12 = 0 $$
For now, let's assume the terms with $x^2$ cancel out, I will discuss this some more later.
Step 6: subtract the pure number part from both sides.
$$4x -12 \ \underline{+ 12} = \underline{+12} $$
Step 7: divide both sides by the coefficient of $x$ and simplify the fractions.
$$ \frac{4x}{4} = \frac{-12}{4} \\ x = - 3$$

applying all these steps to your second equation:
$$6 - 2(x – 4)^2 = 5 – 2(x + 3)(x – 1) \\ \text{apply Step 2 in reverse} \\  6 - 2 (x-4)(x-4) = 5 - 2 (x+3)(x-1) \\ \text{apply Step 1} \\ 6 - 2(x(x-4) - 4(x-4))=5-2(x(x-1) + 3(x-1)) \\ 6 - 2(xx - 4x - 4x + 16) = 5 - 2(xx - x + 3x - 3) \\ \text{apply Step 2} \\ 6 -2(x^2 - 4x - 4x + 16) = 5 - 2(x^2 - x + 3x - 3) \\ \text{apply Step 3} \\ 6 - 2x^2 + 8x + 8x - 32 = 5 - 2x^2 + 2x - 6x + 6 \\ \text{apply Step 4} \\ 6 - 2x^2 + 8x + 8x - 32 - 5 + 2x^2 - 2x + 6x - 6 = 0 \\ \text{apply Step 5} \\ 20x - 37 = 0 \\ \text{apply Step 6} \\ 20x = 37 \\ \text{apply Step 7} \\ x = \frac{37}{20}$$
Notice how we had to apply the second step in reverse before the first one, and again after. It is important to remember that math is not a set of recipes to learn by heart, sometimes you will need to look at the problem and decide what the best strategy to solve it is. For example, in the previous explanation, I could have added the like terms "$-2x -6x$" as soon as they appeared in the parentheses at the end of step 1, and saved some time copying them both from line to line. I also could have simplified the $-x^2$ from both sides as soon as it was out of the parentheses at the end of step 3.

Now let's discuss the assumption I made earlier: I said that we can assume the $x^2$ will cancel out. This is not always true, but to solve the equations where it's not we need another tool.

Let's say that your final result at the end of step 5 is of the type $ax^2 + bx + c = 0$, where $a,b,c$ are pure numbers. Let's also divide both sides by $a$ to get $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$. for simplicity, let $p := \frac b a$ and $q := \frac c a$.
Since we know that a product of two values is 0 if and only if at least one of the values is 0, we can try to rewrite the left hand side of our equation as the product of two binomials of degree 1: so we are looking for numbers $r, s$ such that
$$(x + r)(x + s) = x^2 + px + q$$.
by applying the distributive property twice on the left hand side, we get
$$x^2 + (r+s) x + rs = x^2 + px + q \\ \text {which is only true if} \\ p = r + s,\quad q = rs$$
For example, given the equation $x^2 + 3x + 2 = 0$, we can guess the values $r = 2, s = 1$ since $p = 3 = 1 + 2, q = 2 = 1 \cdot 2$. Our equation will now be: $(x+r)(x+s) = 0$, which as discussed before is true if and only if $x+r = 0$, or $x+s = 0$, or both. Now we can apply step 7 of our previous process from before to get the two solutions: $x = -r$, or $x = -s$.

In general, there is a formula that will give you the values of $r$ and $s$ without guessing, but it involves a square root: given the equation $ax^2 + bx + c = 0$
$$ -r,  -s = \lambda_1, \lambda_2 = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$$
(the "$\pm$" symbol means that one solution has the "$+$" sign there, while the other has the "$-$" sign).

This reveals a test you can make before you guess: since the "$\sqrt \ $" operation (on the real numbers) is only defined on positive or zero numbers, if the quantity $\Delta = b^2 - 4ac < 0$ then you can say that there are no real solutions to the equation.