Solve (2x+3)(2x-3)=7: Binomial Multiplication Equation

Let's solve the equation. First, we'll simplify the algebraic expressions using the abbreviated multiplication formula for difference of squares:

$(a+b)(a-b)=a^2-b^2$

We will then apply the mentioned rule and open the parentheses in the expression in the equation:

$(2x+3)(2x-3)=7 \\ (2x)^2-3^2=7 \\ 4x^2-3^2=7$

In the final stage, we distributed the exponent over the parentheses to both multiplication terms inside the parentheses, according to the laws of exponents:

$(ab)^n=a^nb^n$

Let's continue and combine like terms, by moving terms:

$4x^2-3^2=7 \ 4x^2-9=7 \\ 4x^2-16=0$

Next - we can see that the equation is of second degree and that the coefficient of the first-degree term is 0, so we'll try to solve it using repeated use (in reverse) of the abbreviated multiplication formula for the difference of squares mentioned earlier:

$4x^2-16=0\stackrel{?}{\leftrightarrow } a^2-b^2\\ \downarrow\\ (2x)^2-4^2=0 \stackrel{!}{\leftrightarrow } a^2-b^2\\ \downarrow\\ \boxed{(2x+4)(2x-4)=0} \leftrightarrow (a+b)(a-b)\\$

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplying expressions equals zero,

Therefore we get two simple equations and we'll solve them by isolating the unknown in each:

$2x+4=0\\ 2x=-4\hspace{8pt}\text{/}:2\\ \boxed{x=-2}$

or:

$2x-4=0\\ 2x=4\hspace{8pt}\text{/}:2\\ \boxed{x=2}$

Let's summarize the solution to the equation:

$(2x+3)(2x-3)=7 \\ \downarrow\\ (2x)^2-3^2=7 \\ 4x^2-16=0\\ \downarrow\\ (2x+4)(2x-4)=0\\ \downarrow\\ 2x+4=0\rightarrow\boxed{x=-2}\\ 2x-4=0\rightarrow\boxed{x=2}\\ \downarrow\\ \boxed{x=2,-2}$

Therefore the correct answer is answer B.