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Quadratic equations are polynomial equations of degree 2. They take the general form:

• \)

Here, a, b, and c are constants, with a not equal to zero. The term ²¹³æ² is the quadratic term, bx is the linear term, and c is the constant term. Since they involve squared terms, these equations frequently model situations involving area or any scenario that follows a path described by a parabola.
Solving quadratic equations can be done through various methods like factorization, completing the square, or using the quadratic formula. In our exercise, we initially face an equation that requires simplification using variable substitution to transform it into a clearer quadratic format.
This ability to recognize quadratic forms in disguised equations is crucial for simplifying the approach to solution.

The quadratic formula is a fundamental method for solving any quadratic equation. It is especially useful when factorization is challenging. The formula is written as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

With this formula, a, b, and c are coefficients from the quadratic equation ²¹³æ² + bx + c = 0. The expression under the square root, \(b^2 - 4ac\), is known as the discriminant, determining the nature and number of roots. A positive discriminant indicates two distinct real roots, zero indicates one real root, and negative implies no real roots, as solutions are complex.
In our example, after simplifying using substitution, the application of the quadratic formula aids in finding the solutions for the new variable.Step-by-step application ensures accurate results, transforming the math from abstract to concrete.

Variable substitution is a powerful technique in algebra to simplify complex equations. In the exercise at hand, we use this method by introducing a new variable, u. This reduces the complexity of the original equation, making it easier to work with.
Through the substitution of u = x - 5, the transformed quadratic equation \(u^2 - 4u - 21 = 0\) emerges. This reframing allows us to apply the quadratic formula more straightforwardly than directly solving for x.

• Identify a recurring binomial or expression.
• Substitute it with a symbol like u.
• Solve the resultant simpler equation.
• Re-substitute to solve for the original variable.

This technique is particularly useful when dealing with nested or intricate algebraic expressions, turning a daunting task into a manageable one.