The really important thing to remember here is that we can't "distribute" a square over addition. When we see that little exponent of 2, we need to think of that as telling us to multiply $x+9$ by itself: $$(x+9)^2 = (x+9)(x+9).$$ From there we use distribution (aka FOIL) to expand: $$(x+9)(x+9) = x\cdot x + 9x + 9x + 9\cdot 9$$ $$=x^2 + 18x + 81$$

We could use our "$ac$" method here, or we could maybe recognize a special pattern. Notice that 14 is $7\cdot 2$ and 49 is $7^2$; this is a perfect square trinomial! Therefore, we can factor it as follows: $$x^2 - 14x + 49 = (x-7)(x-7) = (x-7)^2.$$

Here, we first notice that 5, 40, and 80 have a common factor of 5. Let's factor that out first and see where we can go from there.

$$5x^2 + 40x + 80 = 5(x^2 + 8x + 16)$$

Now, we see again that 8 is $4\cdot 2$ and 16 is $4^2$, so $x^2 + 8x + 16$ is another perfect square trinomial. Therefore, our final factorization is $$5(x+8)^2$$

Our first goal is to identify what our $n$ should be. We only have two terms, $x^2$ and $6x$. When we match this up to our binomial squares pattern, we see $$\begin{array}{ccc} x^2 & + \textcolor{Maroon}{6}x & \\ x^2 & +\textcolor{Maroon}{2n}x& +\textcolor{Blue}{n^2}\\ \end{array}$$

When we line up our terms, we see that our $2n$ and $6$ match up. If we let $2n = 6$, we see that $n$ needs to be 3. That makes the missing $n^2$ term $3^2 = 9$.

However, we can't just add a 9 onto the end of our expression, because that would completely change it! The expression $x^2 + 6x$ is different from $x^2 + 6x + 9$. What we're always allowed to add, though, is a fancy form of zero. So, we'll add $+9-9$ to the end of our expression.

$$x^2 + 6x = x^2 + 6x + 9 - 9$$

Now, we'll factor the perfect square portion of our expression, and leave the extra -9 at the end!

$$x^2 + 6x = (x+3)^2 - 9$$

Let's line things up again.

$$\begin{array}{ccc} x^2 & \textcolor{Maroon}{-26}x &\\ x^2 & \textcolor{Maroon}{+2n}n & +\textcolor{Blue}{n^2}\\ \end{array}$$

Here we see $2n = -26$, so $n = -13$. That makes $n^2 = (-13)^2 = 169$, so we'll add $169-169$ (our fancy form of zero) to the end of our expression, then factor the perfect square part, leaving the -169 at the end.

$$x^2 -26x + 169 - 169$$ $$ =(x-13)^2 - 169$$

Let's line things up again.

$$\begin{array}{ccc} x^2 & \textcolor{Maroon}{-9}x &\\ x^2 & \textcolor{Maroon}{+2n}n & +\textcolor{Blue}{n^2}\\ \end{array}$$

Here we see $2n = -9$, so $n = -\frac92$. It's not pretty, but that makes $n^2 = \left(-\frac92\right)^2 = \frac{81}4$, so we'll add $\frac{81}4-\frac{81}4$ (our fancy form of zero) to the end of our expression, then factor the perfect square part, leaving the $-\frac{81}4$ at the end.

$$x^2 -9x + \frac{81}4-\frac{81}4$$

Now, the factoring seems especially tricky, but remember the pattern we're using! We want to end up with $(x+n)^2$ for the factored portion, and we already figured out that $n = -\frac92$. So, our final factored answer should be $$\left(x-\frac92\right)^2 -\frac{81}4.$$

First, we'll work on completing the square for the $x^2 + 6x$ terms. Using previous examples, we'll see that $n = 3$, so $n^2 = 9$, and we'll have: $$\begin{align*} x^2 + 6x &= 40\\ x^2 + 6x + 9-9 &= 40\\ (x+3)^2 - 9 &= 40\\ \end{align*}$$ Now this equation is nicely set up to use the square root strategy, if we move the 9 to the other side. $$\begin{align*} (x+3)^2 &= 49\\ \sqrt{(x+3)^2} &= \sqrt{49}\\ x+3 &= \pm 7\\ \end{align*}$$ We finish this up by solving $x+3 = 7$ and $x + 3 = -7$, for solutions of $x = 4$ and $x = -10$, respectively.

First, we'll work on completing the square for the $x^2 + 8x$ terms. Using previous examples, we'll see that $n = 4$, so $n^2 = 16$, and we'll have: $$\begin{align*} x^2 + 8x &= 48\\ x^2 + 8x + 16-16 &= 48\\ (x+4)^2 - 16 &= 48\\ \end{align*}$$ Now this equation is nicely set up to use the square root strategy, if we move the 16 to the other side. $$\begin{align*} (x+4)^2 &= 64\\ \sqrt{(x+4)^2} &= \sqrt{64}\\ x+4 &= \pm 8\\ \end{align*}$$ We finish this up by solving $x+4 = 8$ and $x + 4 = -8$, for solutions of $x = 4$ and $x = -12$, respectively.

Before we can make use of our $(x+n)^2$ pattern, we need to factor out the leading 3: $$3x^2 - 12x = 3(x^2 - 4x)$$

Now, we're going to match the $x^2 -4x$ portion of our expression up with the $x^2 + 2nx + n^2$ pattern, but we're going to keep that factored 3 in mind as we do.

$$\begin{array}{rccc} 3 & (x^2 & \textcolor{Maroon}{-4}x & ~~)\\ & (x^2 & \textcolor{Maroon}{+2n}x & +\textcolor{Blue}{n^2})\\ \end{array}$$

Now we see that $2n = -4$, so $n = -2$. Here is where we need to be careful. Our $n^2$ is still $(-2)^2 = 4$, but when we add our fancy zero, $4-4$, that's all inside our parentheses, being multiplied by a 3 on the outside: $$3(x^2 -4x + 4 -4).$$

Now, we'll factor the perfect square portion, and see what we have left.

$$3((x-2)^2 -4)$$

If we distribute the 3, we have $$3(x-2)^2 - 3\cdot 4 = 3(x-2)^2 -12$$ for our final answer.

Before we can make use of our $(x+n)^2$ pattern, we need to factor out the leading 4: $$4x^2 - 24x = 4(x^2 - 6x)$$

Now, we're going to match the $x^2 -6x$ portion of our expression up with the $x^2 + 2nx + n^2$ pattern, but we're going to keep that factored 4 in mind as we do.

$$\begin{array}{rccc} 4 & (x^2 & \textcolor{Maroon}{-6}x & ~~)\\ & (x^2 & \textcolor{Maroon}{+2n}x & +\textcolor{Blue}{n^2})\\ \end{array}$$

Now we see that $2n = -6$, so $n = -3$. Here is where we need to be careful. Our $n^2$ is still $(-3)^2 = 9$, but when we add our fancy zero, $9-9$, that's all inside our parentheses, being multiplied by a 4 on the outside: $$4(x^2 -6x + 9-9).$$

Now, we'll factor the perfect square portion, and see what we have left.

$$4((x-3)^2 -9)$$

If we distribute the 4, we have $$4(x-3)^2 - 9\cdot 4 = 4(x-3)^2 -36$$ for our final answer.

Start by factoring out the 2 from only the first terms. Remember, the 14 isn't really involved in completing the square; we're only concerned about the $ax^2$ and $bx$ terms.

$$2x^2 + 16x + 14 = 0$$ $$2(x^2 + 8x + ~~~) + 14 = 0$$

Now, we want to complete the square for $x^2 + 8x$. We see that $2n = 8$, so $n = 4$ and $n^2 = 16$. Then we add $16 -16$ inside our parentheses.

$$2(x^2 + 8x + 16 - 16) + 14 = 0$$

Factor, then simplify, and we have $$2((x+4)^2 -16) + 14 = 0$$ $$2(x+4)^2 - 2\cdot 16 + 14 = 0$$ $$2(x+4)^2 -18 = 0$$

This equation is ready for solving via square roots. We'll move the 18 to the other side of the equals sign, then divide by 2, then root both sides to undo the square.

$$2(x+4)^2 = 18$$ $$(x+4)^2 = 9$$ $$\sqrt{(x+4)^2} = \sqrt{9}$$ $$x+4 = 9 \text{ or } x + 4 = -9$$ $$x = 5 \text{ or } x = -13$$

Start by factoring out the 2 from the terms on the left hand side. This feels a little weird, since 2 isn't a factor of 3. Keep in mind that factoring out a common factor is kind of like dividing each term by that factor.

$$2x^2 -3x = 20$$ $$2(x^2 - \frac32 x + ~~~) = 20$$

Now, we want to complete the square for $x^2 -\frac32 x$. We see that $2n = -\frac32$, so $n = -\frac34$ and $n^2 = \frac9{16}$. Then we add $\frac9{16}-\frac9{16}$ inside our parentheses.

$$2(x^2 -\frac32x +\frac9{16}-\frac9{16}) = 20$$

Factor (remembering that our $n=\frac34$, then simplify, and we have $$2((x-\frac34)^2 -\frac9{16})= 20$$ $$2(x-\frac34)^2 - 2\cdot \frac9{16} = 20$$ $$2(x-\frac34)^2 - \frac98 = 20$$

This equation is ready for solving via square roots. We'll move the $\frac98$ to the other side of the equals sign, then divide by 2, then root both sides to undo the square.

$$2(x-\frac34)^2 = \frac{169}{8}$$ $$(x-\frac34)^2 = \frac{169}{16}$$ $$\sqrt{(x-\frac34)^2} = \sqrt{\frac{169}{16}}$$ $$x-\frac34 = \frac{13}{4} \text{ or } x - \frac34 = -\frac{13}{4}$$ $$x = \frac{16}{4} \text{ or } x = -\frac{10}{4}$$ $$x=4, -\frac52$$