1. **State the problem:** Factorize the quadratic equation $6t^2 - 16t + 10 = 0$. 2. **Recall the factoring formula:** For a quadratic $at^2 + bt + c = 0$, we look for two numbers that multiply to $a \times c$ and add to $b$. 3. **Calculate product and sum:** Here, $a=6$, $b=-16$, $c=10$. So, product = $6 \times 10 = 60$, sum = $-16$. 4. **Find two numbers:** The numbers that multiply to 60 and add to -16 are -10 and -6. 5. **Rewrite middle term:** $6t^2 - 10t - 6t + 10 = 0$. 6. **Group terms:** $(6t^2 - 10t) + (-6t + 10) = 0$. 7. **Factor each group:** $2t(3t - 5) - 2(3t - 5) = 0$. 8. **Factor out common binomial:** $(2t - 2)(3t - 5) = 0$. 9. **Simplify first factor:** $2(t - 1)(3t - 5) = 0$. 10. **Final factorized form:** $(t - 1)(3t - 5) = 0$. Thus, the factorization of $6t^2 - 16t + 10$ is $(t - 1)(3t - 5)$.