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The quadratic formula is a powerful tool for solving quadratic equations. Quadratic equations are those that include a term with the variable squared, looking like this: \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are known constants, and \( x \) is the variable we want to solve for. The quadratic formula is given as:

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

This formula allows you to find the values of \( x \) that make the equation true. It encompasses all possible solutions for any quadratic equation, whether the roots are real or complex. To apply the formula, plug in values for \( a \), \( b \), and \( c \), and perform the arithmetic. The "plus-minus" (\( \pm \)) sign indicates that there could be two possible solutions. Don't forget to check your calculations with the discriminant \( b^2 - 4ac \). This part under the square root determines the nature of the roots:

• If it’s positive, there are two distinct real roots.
• If zero, there’s exactly one real root.
• If negative, the roots are complex.

In physics, the equation related to projectiles often requires using quadratic principles. The equation \( s = -16t^2 + v_0 t \) is derived from the laws of kinematics and describes the vertical motion of a projectile under gravity in cases where air resistance is negligible. In this equation:

• \( s \) represents the vertical position or height.
• \( t \) is the time the projectile has been in motion.
• \( v_0 \) is the initial velocity of the projectile.
• The term \(-16t^2\) represents the gravitational pull affecting the motion, with \(-16\) approximating the acceleration due to gravity (in feet per second squared).

By solving for \( t \) in the expression \( s = -16t^2 + v_0 t \), you can determine how long it takes for a projectile to reach a certain height. This is particularly useful in real-world applications, such as determining when a ball thrown into the air will hit the ground. The equation is quadratic by nature because of the \( t^2 \) term, requiring the quadratic formula for solving.

Simplifying expressions is a crucial step in solving equations to make them more manageable. By simplifying, you reduce complexity and make calculations easier to handle. Let's consider the expression from the solution process: \( t = \frac{-v_{0} \pm \sqrt{v_{0}^2 + 64s}}{32} \).When you simplify an expression:

• Combine like terms to shorten the expression.
• Factor out common elements if possible, to reduce computation.
• Focus on reducing fractions by canceling out common factors from the numerator and denominator.

The given final simplification is achieved by factoring out common terms from the quadratic formula application, resulting in a cleaner and simpler expression. For instance, if a common factor like \( 32 \) appears in both the numerator and denominator in a fraction, you can divide them to simplify. Always ensure your expressions after simplifying maintain equality with the original equation.