Solve the Quadratic Equation: 16a² + 20a + 20 = -5 - 20a

Let's solve the given equation:

$16a^2+20a+20=-5-20a$

First, let's organize the equation by moving terms and combining like terms:

$16a^2+20a+20=-5-20a \\ 16a^2+20a+20+5+20a =0\\ 16a^2+40a+25=0$

Note that we are able to factor the expression on the left side by using the perfect square trinomial formula for a binomial squared:

$(\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+2\textcolor{red}{x}\textcolor{blue}{y}+\textcolor{blue}{y}^2$

As shown below:

$16=4^2\\ 25=5^2$

Using the law of exponents for powers applied to products in parentheses (in reverse):

$x^ny^n=(xy)^n$

Therefore, first we'll express the outer terms as a product of squared terms:

$16a^2+40a+25=0 \\ 4^2a^2+40a+5^2=0 \\ \downarrow\\ (\textcolor{red}{4a})^2+40a+\textcolor{blue}{5}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+\underline{2\textcolor{red}{x}\textcolor{blue}{y}}+\textcolor{blue}{y}^2$

And the expression on the left side of the equation that we obtained in the last step:

$(\textcolor{red}{4a})^2+\underline{40a}+\textcolor{blue}{5}^2=0$

Note that the terms $(\textcolor{red}{4a})^2,\hspace{6pt}\textcolor{blue}{5}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a single line):

$(\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+\underline{2\textcolor{red}{x}\textcolor{blue}{y}}+\textcolor{blue}{y}^2$

In other words - we are querying whether we can express the expression on the left side of the equation as:

$(\textcolor{red}{4a})^2+\underline{40a}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ (\textcolor{red}{4a})^2+\underline{2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0$

And indeed it holds that:

$2\cdot4a\cdot5=40a$

Therefore, we can express the expression on the left side of the equation as a perfect square binomial:

$(\textcolor{red}{4a})^2+2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0\\ \downarrow\\ (\textcolor{red}{4a}+\textcolor{blue}{5})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable and dividing both sides of the equation by the variable's coefficient:

$(4a+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ 4a+5=\pm0\\ 4a+5=0\\ 4a=-5\hspace{8pt}\text{/}:4\\ \boxed{a=-\frac{5}{4}}$

Let's summarize the solution of the equation:

$16a^2+20a+20=-5-20a \\ 16a^2+40a+25=0 \\ \downarrow\\ (\textcolor{red}{4a})^2+2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0\\ \downarrow\\ (\textcolor{red}{4a}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ 4a+5=0\\ \downarrow\\ \boxed{a=-\frac{5}{4}}$

Therefore the correct answer is answer D.