To solve the equation \( x = vt + at^2 \) for \( t \), we need to rewrite it as a standard quadratic equation. Here's the step-by-step process:

1. Start with the equation:
   \[
   x = vt + at^2
   \]

2. Rearrange it in the standard quadratic form \( at^2 + bt + c = 0 \):
   \[
   at^2 + vt - x = 0
   \]
   Here, \( a \) is the coefficient of \( t^2 \), \( v \) is the coefficient of \( t \), and \( -x \) is the constant term.

3. Now, apply the quadratic formula:
   \[
   t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   In our case, \( a = a \), \( b = v \), and \( c = -x \).

4. Substituting the values:
   \[
   t = \frac{-v \pm \sqrt{v^2 - 4a(-x)}}{2a}
   \]
   Simplifying the expression:
   \[
   t = \frac{-v \pm \sqrt{v^2 + 4ax}}{2a}
   \]

Thus, the solution for \( t \) is:
\[
t = \frac{-v \pm \sqrt{v^2 + 4ax}}{2a}
\]

There are two possible solutions for \( t \), depending on the \( \pm \) sign.

Would you like more details or have any questions?

Here are 5 related questions:
1. How does the quadratic formula work for general equations?
2. What are the physical interpretations of the terms \( v \), \( a \), and \( x \) in this context?
3. Can you derive a solution if \( a = 0 \)?
4. What happens if the discriminant \( v^2 + 4ax \) is negative?
5. How can we apply this formula in real-world motion problems?

**Tip:** Always check the discriminant \( b^2 - 4ac \) when solving quadratic equations to determine the nature of the solutions.

Learn how to solve the kinematic equation x = vt + at² for time (t) using the quadratic formula. This detailed guide walks you through algebraic manipulation and applying the quadratic formula to solve for t. Ideal for students in Grades 9-12.

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