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When tackling problems involving quadratic equations, the distributive property is a powerful and necessary tool. This property allows us to multiply a single term by each term in a binomial or polynomial expression. Think of it like distributing objects fairly among children at a party. In terms of algebra, we distribute a number or variable outside the parenthesis to every term inside the parenthesis.

In the exercise, we first applied the distributive property to expand \( (x-2)(x+3) \), meaning we wanted to multiply each term in \(x-2\) by each term in \(x+3\). This looks like:

• Multiply \(x\) by \(x+3\)
• Then multiply \(-2\) by \(x+3\)

By consistently applying the distributive property like this, expressions become simplified, setting a firm foundation for solving the equation.

Expanding expressions is the process of eliminating the parentheses in an algebraic expression. This step is essential for simplifying complex equations and forms a core strategy when dealing with quadratic equations.

In our original problem, after using the distributive property, we expanded the expression \( (x-2)(x+3) \) to \( x^2 + x - 6 \). Here is how expanding helps:

• It changes a product of sums or differences into a sum or difference of products, which is often more manageable.
• Expanding makes it easier to combine like terms, further simplifying the process.

By expanding correctly, you can transform complex expressions into ones that are easier to handle and solve, bringing you one step closer to finding the solution.

The quadratic formula is a trusty companion for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). When factoring is not easy or possible, the quadratic formula steps in to solve for the variable.
To apply this formula, simply plug the coefficients (

• \(a\) - the coefficient of \(x^2\)
• \(b\) - the coefficient of \(x\)
• \(c\) - the constant term

into the formula: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\).
In our exercise, using \(a = 2\), \(b = -5\), and \(c = -3\), we calculated the solutions as \(x = 3\) and \(x = -0.5\). This formula allows us to find solutions effectively, even when equations get tricky.