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The quadratic formula is a powerful tool for solving second-degree polynomial equations of the form \(ax^2 + bx + c = 0\). With the formula, we can find the values of \(x\), even when factoring seems complicated or impossible. The formula itself is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Where:- \(a\) is the coefficient of \(x^2\)- \(b\) is the coefficient of \(x\)- \(c\) is the constant termBy using this formula, you directly calculate the roots of the equation by substituting the coefficients. It is especially useful since it accounts for both real and complex solutions based on the value of the discriminant \(b^2 - 4ac\). Here, the quadratic formula helped us find exact real solutions for the quadratic part of the equation \(6x^2 + 6x - 13 = 0\). This is done by first calculating the discriminant and then solving for \(x\).

Factoring is a method used to break down polynomials into simpler, "factorable" components. It involves finding values of \(x\) that turn the polynomial into a product of simpler terms. In our case, we began by factoring the polynomial \(6x^3 + 6x^2 - 13x = 0\).Here’s how we did it:- First, we factored out the greatest common factor from each term, which was \(x\).- This gave us \(x(6x^2 + 6x - 13) = 0\).- Finding the factor \(x\) immediately offers one solution: \(x = 0\).The remaining quadratic \(6x^2 + 6x - 13\) required the use of the quadratic formula to find additional solutions. Factoring simplifies complex polynomials into smaller parts that can often be solved using known methods like factoring further, completing the square, or leveraging the quadratic formula. It is crucial to recognize the most effective factoring step to employ and use them together with other methods to solve for unknowns in polynomial equations.

Real solutions of an equation are values of \(x\) that satisfy the equation and are real numbers. For polynomial equations, real solutions correspond to where the graph of the polynomial intersects the \(x\)-axis.Here's how we identified the real solutions for the equation \(6x^3 + 6x^2 - 13x = 0\):- By factoring the polynomial, we found one solution as \(x = 0\).- For the quadratic \(6x^2 + 6x - 13 = 0\), we used the quadratic formula.The calculation of the discriminant revealed that it was positive, indicating two distinct real solutions for the quadratic. These solutions were:- \(x = 1.05\)- \(x = -2.05\)Thus, the set of real solutions for the entire equation is \(x = 0\), \(x = 1.05\), and \(x = -2.05\). Real solutions are usually what you graph, look at in practical scenarios, and use in a majority of real-world applications.