When solving quadratic equations, the goal is to find values for \( x \) that satisfy the equation, similar to solving linear equations. However, quadratic equations, which typically take the form \( ax^2 + bx + c = 0 \), often yield two solutions for \( x \). To solve these equations, we can utilize the method of factoring.

To begin, we need to express the quadratic equation in standard form, ensuring all terms are on one side and arranged in descending order of power. For example, if we have the equation \( x^2 - 9x = -20 \), we can add 20 to both sides to rewrite it as \( x^2 - 9x + 20 = 0 \).

Next, we factor the quadratic expression. In this case, we identify the coefficients \( a \), \( b \), and \( c \) from the standard form. Here, \( a = 1 \), \( b = -9 \), and \( c = 20 \). We then apply the \( ac \) method, which involves finding two numbers that multiply to \( ac \) (which is \( 20 \)) and add to \( b \) (which is \( -9 \)). The factors that meet these criteria are \( -4 \) and \( -5 \), since \( -4 \times -5 = 20 \) and \( -4 + -5 = -9 \).

With the factors identified, we can express the quadratic as \( (x - 4)(x - 5) = 0 \). To find the solutions for \( x \), we set each factor equal to zero:

1. \( x - 4 = 0 \) leads to \( x = 4 \)

2. \( x - 5 = 0 \) leads to \( x = 5 \)

Thus, the solutions to the quadratic equation are \( x = 4 \) and \( x = 5 \). To verify these solutions, we can substitute them back into the original equation. For \( x = 4 \):

\( 4^2 - 9(4) + 20 = 0 \) simplifies to \( 16 - 36 + 20 = 0 \), confirming that \( 4 \) is indeed a solution.

Similarly, substituting \( x = 5 \) will also yield a true statement, confirming both values are correct. This process illustrates how factoring can effectively solve quadratic equations, allowing us to find the necessary values for \( x \) that satisfy the equation.