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$4y^2+32y = 0$

$4(y^2+8y+64-64) = 0$

$4(y+4)^2 = 64$

Is that correct?

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    Your first two 64's should be 16's. When you pull the last one out, it becomes a 64. – 2017-01-03
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    This is not a linear algebra question as you can see here https://en.wikipedia.org/wiki/Linear_algebra – 2017-01-03
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    Thanks Everyone, I appreciate the help you guys provide. – 2017-01-03
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    I'm not sure why you don't just divide the original equation through by $4$ to start with. – 2017-01-03
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    @MarkBennet I agree with that as well – 2017-01-03

4 Answers 4

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Let's take a closer look at that first step:

$$4(y^2+8y+64-64)=0$$

Clearly you are able to see the perfect square, but there isn't one. Notice that $(y+4)^2=y^2+8y+16$. So let's change it to $16$.

$$4(y^2+8y+16-16)=0$$

Now we do have a perfect square, but you seem to make a mistake I see happen all the time.

$$\begin{array}{c|c}\text{What you do}&\text{What I do}\\\hline4(y+4)^2-16&4((y+4)^2-16)\end{array}$$

See the big difference some parenthesis make? Now, we should end up with

$$4(y+4)^2-64=0$$

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    Thanks I just wanted to confirm, My math is very rusty. – 2017-01-03
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    @Reboot No problem, we all make mistakes. More often than not – 2017-01-03
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$4y^2+32y = 0$

$4(y^2+8y+16-16) = 0$

$4(y+4)^2 = 64$

$(y+4)^2 = 16$

$y+4= \pm4$

$y_1=-8$, $y_2=0$

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    Thanks I just wanted to confirm, My math is very rusty. – 2017-01-03
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When you expand $(a\pm b)^2$, look at the coefficient of $x$ and the constant.$$(a\pm b)^2=a^2\pm2ab+b^2\tag1$$ Assuming that it's just $a^2$, we see that to complete the square, you have to divide by $2$ and square the result. Or,$$a^2+ba+\left(\dfrac b2\right)^2=\left(a+\dfrac b2\right)^2\tag2$$ Now, what was wrong with your work is that instead of dividing by $2$, you just squared the coefficient. So really, it should be$$4\left(y^2+8y+\left(\dfrac 42\right)^2-\left(\dfrac 42\right)^2\right)=0$$$$4(y^2+8y+16)-64=0$$$$4(y+4)^2=64$$

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I made a comment, and I would do things slightly differently, so $$4y^2+32y=0$$ divide by $4$$$y^2+8y=0$$ add $\left(\frac 82\right)^2=4^2=16$ to both sides $$y^2+8y+16=16$$Rewrite the left-hand side as a square $$(y+4)^2=16$$If you want the $4$ back you can multiply through by $4$ at the end. If you want a pure square this is $(2y+8)^2=64$. The version you have put in your question does not express the left-hand side as a square.