\( 8 v^{2}-21 v-9=0 \) \( 8 v^{2}-\square v+\square v-9=0 \) \( 8 v(v-\square)+3(v-\square=0 \) \( \left(8 v^{2}-\square v\right)+(\square v-9)=0 \) \( 8 v(v-\square)+(v-\square)=0 \) \( (3 v+\square)(v-\square)=0 \) \( 8 v+\square=0 \) or \( v-\square=0 \) \( 8 v=\square \) by \( \quad \square=\square \) \( v=\frac{\square}{8} \) or \( v=\square \)

To solve the quadratic equation \( 8v^2 - 21v - 9 = 0 \), you can factor the expression. A good starting point is finding two numbers that multiply to \( 8 \times -9 = -72 \) and add to \(-21\). Those numbers are \(-24\) and \(3\). This allows you to rewrite the equation as \( 8v^2 - 24v + 3v - 9 = 0 \). From there, group and factor the equation to reach the factored form. Ultimately, you'd find the roots using the zero-product property, which leads to solving for \( v \) in both cases derived from your factors. Using the quadratic formula is a real-world application of this problem, especially in physics. For instance, if you're calculating the trajectory of a thrown object, quadratic equations help you determine the object's height at various points in time, which can result in predicting its landing spot or optimizing angles for maximum distance.