It looks like you're trying to solve a quadratic equation of the form \( y = -\frac{25}{36}x^2 + \frac{50}{3}x \). To determine how far the firework will travel, you would typically set the equation equal to zero and solve for \( x \):
\[ -\frac{25}{36}x^2 + \frac{50}{3}x = 0 \]
Now we can factor out \( x \):
\[ x\left(-\frac{25}{36}x + \frac{50}{3}\right) = 0 \]
This gives us one solution:
\[ x = 0 \]
Now, let's solve the remaining equation:
\[ -\frac{25}{36}x + \frac{50}{3} = 0 \]
To isolate \( x \), we first move the term involving \( x \):
\[ -\frac{25}{36}x = -\frac{50}{3} \]
Now, multiply both sides by \(-1\):
\[ \frac{25}{36}x = \frac{50}{3} \]
To eliminate the fraction, multiply both sides by \( \frac{36}{25} \):
\[ x = \frac{50}{3} \cdot \frac{36}{25} \]
Now perform the multiplication:
\[ x = \frac{50 \times 36}{3 \times 25} = \frac{1800}{75} = 24 \]
Thus, the solutions to the equation are \( x = 0 \) and \( x = 24 \). The firework will reach the ground at \( x = 24 \). So, from the values in your table, if \( 24 \) is provided, it is a solution to this equation.