The expression you've provided is $(7s + 2t)^2$. To simplify this, we apply the square of a binomial formula, which states:

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

Here, $a = 7s$ and $b = 2t$. Applying the formula:

$(7s + 2t)^2 = (7s)^2 + 2(7s)(2t) + (2t)^2$

Now, let's calculate each term:

1. $(7s)^2 = 49s^2$
2. $2(7s)(2t) = 28st$
3. $(2t)^2 = 4t^2$

Thus, the expanded form of the expression is:

$(7s + 2t)^2 = 49s^2 + 28st + 4t^2$

Would you like more details on the binomial expansion process or have any further questions?

Here are 5 questions related to this problem:

1. How do we generalize the expansion of higher powers, like $(7s + 2t)^3$?
2. How does this method change if there are negative signs, such as $(7s - 2t)^2$?
3. Can you expand other binomials, like $(5x + 3y)^2$, similarly?
4. How does this concept relate to factoring quadratics?
5. Can you use this expansion in solving quadratic equations?

Tip: Remember that the square of a binomial always results in three terms: a square, a cross-product, and another square!