Find the product of the roots of the equation $18t^2 + 45t -500 =0$.
The product of the roots equals the constant term divided by the coefficient of the quadratic term, or $(-500)/18 = \boxed{-\frac{250}{9}}$.
-\frac{250}{9}
Level 4
Algebra
Problem: Find the product of the roots of the equation $18t^2 + 45t -500 =0$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the product of the roots of a quadratic equation, we can use Vieta's formulas. For a quadratic equation of the form $at^2 + bt + c = 0$, the product of the roots is given by the formula $\frac{c}{a}$. 

In this case, the coefficients are $a = 18$, $b = 45$, and $c = -500$. 

First, we identify the values of $a$ and $c$:
- $a = 18$
- $c = -500$

Next, we apply Vieta's formula to find the product of the roots:
Product of the roots = $\frac{c}{a} = \frac{-500}{18}$.

Now, we simplify the fraction:
- The numerator is $-500$.
- The denominator is $18$.

To simplify $\frac{-500}{18}$, we can divide both the numerator and the denominator by their greatest common divisor, which is $2$:
- $-500 \div 2 = -250$
- $18 \div 2 = 9$

Thus, we have:
Product of the roots = $\frac{-250}{9}$.

This leads us to the final answer, which is $-\frac{250}{9}$.<|END_OF_LATENT|> The final answer is $-\frac{250}{9}$.