Let's solve the given equation:

$60-16y+y^2=-4$First, **let's arrange the equation** by moving terms:

$60-16y+y^2=-4 \\
60-16y+y^2+4=0 \\
y^2-16y+64=0$Now, let's note that we can break down the expression on the left side using the **short multiplication formula for the difference of two squares**:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$This is done using the fact that:

$64=8^2$So let's present the outer term on the right as a square:

$y^2-16y+64=0 \\
\downarrow\\
\textcolor{red}{y}^2-16y+\textcolor{blue}{8}^2=0$**Now let's examine again the short multiplication formula we mentioned earlier:**

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$And the expression on the left side of the equation we got in the last step:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0$Let's note that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2$indeed match the form of the first and third terms in the short multiplication formula (which are highlighted in red and blue),

But in order for us to break down the relevant expression (which is on the left side of the equation) using the short multiplication formula we mentioned, __the match to the short multiplication formula must also occur from the remaining term__, meaning the middle term in the expression (highlighted with __a bottom line__):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$In other words - we'll ask if it's possible to present the expression on the left side of the equation as:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2 =0 \\
\updownarrow\text{?}\\
\textcolor{red}{y}^2-\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}}+\textcolor{blue}{8}^2 =0$__And indeed it holds__ that:

$2\cdot y\cdot8=16y$**So we can present the expression on the left side of the given equation as a difference of two squares**:

$\textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\
\downarrow\\
(\textcolor{red}{y}-\textcolor{blue}{8})^2=0$From here we can take out square roots for the two sides of the equation **(and not forget that there are two possibilities - positive and negative when taking out square roots from an even order equation)**, we'll solve it easily by isolating the unknown in the side:

$(y-8)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\
y-8=\pm0\\
y-8=0\\
\boxed{y=8}$

**Let's summarize** then the solution of the equation:

$60-16y+y^2=-4 \\
y^2-16y+64=0 \\
\downarrow\\
\textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\
\downarrow\\
(\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\
\downarrow\\
y-8=0\\
\downarrow\\
\boxed{y=8}$

__So the correct answer is answer a.__